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Class 11 NCERT Solutions for chemistry chapter 1 Some Basic Concepts of Chemistry
CBSE Class 11 Chemistry chapter 1 Some Basic Concepts of Chemistry is one of the key chapters and consists of important topics that are often asked in class 11 chemistry examination. The chapter deals with topics like the importance of chemistry, atomic mass, molecular mass, different laws and theories like Dalton’s atomic theory, Avogadro Law, the law of conservation of mass etc.
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Subtopics of Chemistry chapter 1 Some Basic Concepts of Chemistry
- Importance Of Chemistry
- Nature Of Matter
- Properties Of Matter And Their Measurement
- The International System Of Units (Si)
- Mass And Weight
- Uncertainty In Measurement
- Scientific Notation
- Significant Figures
- Dimensional Analysis
- Laws Of Chemical Combinations
- Law Of Conservation Of Mass
- Law Of Definite Proportions
- Law Of Multiple Proportions
- Gay Lussac’s Law Of Gaseous Volumes
- Avogadro Law
- Dalton’s Atomic Theory
- Atomic And Molecular Masses
- Atomic Mass
- Average Atomic Mass
- Molecular Mass
- Formula Mass
- Mole Concept And Molar Masses
- Percentage Composition
- Empirical Formula For Molecular Formula
- Stoichiometry And Stoichiometric Calculations
- Limiting Reagent
- Reactions In Solutions.
Class 11 Physics Chapter 8 Gravitation Important questions
Exercise
Q1. Find out the value of molecular weight of the given compounds:
(i) \(CH_{4}\) (ii)\(H_{2}O\) (iii)\(CO_{2}\)
Ans.
(i)\(CH_{4}\) :
Molecular weight of methane, \(CH_{4}\)
= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen)
= [1(12.011 u) +4 (1.008u)]
= 12.011u + 4.032 u
= 16.043 u
(ii) \(H_{2}O\) :
Molecular weight of water, \(H_{2}O\)
= (2 x Atomic weight of hydrogen) + (1 x Atomic weight of oxygen)
= [2(1.0084) + 1(16.00 u)]
= 2.016 u +16.00 u
= 18.016u
So approximately
= 18.02 u
(iii) \(CO_{2}\) :
= Molecular weight of carbon dioxide, \(CO_{2}\)
= (1 x Atomic weight of carbon) + (2 x Atomic weight of oxygen)
= [1(12.011 u) + 2(16.00 u)]
= 12.011 u +32.00 u
= 44.011 u
So approximately
= 44.01u
Q2. Sodium Sulphate (\(Na_{2}SO_{4}\)) has various elements, find out the mass percentage of each element.
Ans.
Now for \(Na_{2}SO_{4}\).
Molar mass of \(Na_{2}SO_{4}\)
= [(2 x 23.0) + (32.066) + 4(16.00)]
=142.066 g
Formula to calculate mass percent of an element = \(\frac{Mass\;of\;that\;element\;in\;the\;compound}{Molar\;mass\;of\;the\;compound}\times 100\)
Therefore, Mass percent of the sodium element:
= \(\frac{46.0g}{142.066g}\times 100\)
= 32.379
=32.4%
Mass percent of the sulphur element:
=\(\frac{32.066g}{142.066g}\times 100\)
= 22.57
=22.6%
Mass percent of the oxygen element:
= \(\frac{64.0g}{142.066g}\times 100\)
=45.049
=45.05%
Q3. Find out the empirical formula of an oxide of iron having 69.9% Fe and 30.1% O_{2} by mass.
Ans.
Percent of Fe by mass = 69.9 % [As given above]
Percent of O_{2} by mass = 30.1 % [As given above]
Relative moles of Fe in iron oxide:
= \(\frac{percent\;of\;iron\;by\;mass}{Atomic\;mass\;of\;iron}\)
= \(\frac{69.9}{55.85}\)
= 1.25
Relative moles of O in iron oxide:
= \(\frac{percent\;of\;oxygen\;by\;mass}{Atomic\;mass\;of\;oxygen}\)
= \(\frac{30.1}{16.00}\)
= 1.88
Simplest molar ratio of Fe to O:
= 1.25: 1.88
= 1: 1.5
\(\approx\) 2: 3Therefore, empirical formula of iron oxide is \(Fe_{2}O_{3}\).
Q4. Find out the amount of CO_{2} that can be produced when
(i) 1 mole carbon is burnt in air.
(ii) 1 mole carbon is burnt in 16 g of O_{2}.
(iii) 2 moles carbon are burnt in 16 g O_{2}.
Ans.
(i) 1 mole of carbon is burnt in air.
\(C+O_{2}\rightarrow CO_{2}\)1 mole of carbon reacts with 1 mole of O_{2} to form one mole of CO_{2}.
Amount of \(CO_{2}\) produced = 44 g
(ii) 1 mole of carbon is burnt in 16 g of O_{2}.
1 mole of carbon burnt in 32 grams of O_{2} it forms 44 grams of \(CO_{2}\).
Therefore, 16 grams of O_{2} will form \(\frac{44\times 16}{32}\)
= 22 grams of \(CO_{2}\)
(iii) 2 moles of carbon are burnt in 16 g of O_{2}.
If 1 mole of carbon are burnt in 16grams of O_{2} it forms 22 grams of \(CO_{2}\)
Therefore, if 2 moles of carbon are burnt it will form
= \(\frac{2\times 22}{1}\)
= 44g of \(CO_{2}\)
Q5. Find out the mass of \(CH_{3}COONa\)(sodium acetate) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of \(CH_{3}COONa\) is \(82.0245\;g\;mol^{-1}\)
Ans.
0.375 Maqueous solution of \(CH_{3}COONa\)
= 1000 mL of solution containing 0.375 moles of \(CH_{3}COONa\)
Therefore, no. of moles of \(CH_{3}COONa\) in 500 mL
= \(\frac{0.375}{1000}\times 1000\)
= 0.1875 mole
Molar mass of sodium acetate = \(82.0245\;g\;mol^{-1}\)
Therefore, mass that is required of \(CH_{3}COONa\)
= \((82.0245\;g\;mol^{-1})(0.1875\;mole)\)
= 15.38 gram
Q6. A sample of HNO_{3} has a density of \(1.41\;g\;mL^{-1}\) find the concentration of HNO_{3} in moles per litre and the mass percent of HNO_{3} in it is 69%.
Ans.
Mass percent of HNO_{3} in sample is 69 %
Thus, 100 g of HNO_{3} contains 69 g of HNO_{3} by mass.
Molar mass of HNO_{3}
= { 1 + 14 + 3(16)} \(g\;mol^{-1}\)
= 1 + 14 + 48
\(= 63g\;mol^{-1}\)
Now, No. of moles in 69 g of \(HNO_{3}\):
= \(\frac{69\:g}{63\:g\:mol^{-1}}\)
= 1.095 mol
Volume of 100g HNO_{3} solution
= \(\frac{Mass\;of\;solution}{density\;of\;solution}\)
= \(\frac{100g}{1.41g\;mL^{-1}}\)
= 70.92mL
= \(70.92\times 10^{-3}\;L\)
Concentration of HNO_{3}
= \(\frac{1.095\:mole}{70.92\times 10^{-3}L}\)
= 15.44mol/L
Therefore, Concentration of HNO_{3} = 15.44 mol/L
Q7. How much Cu (Copper) can be obtained from 100 gram of \(CuSO_{4}\)( copper sulphate)?
Ans.
1 mole of \(CuSO_{4}\) contains 1 mole of Cu.
Molar mass of \(CuSO_{4}\)
= (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00
= 159.5 gram
159.5 gram of \(CuSO_{4}\) contains 63.5 gram of Cu.
Therefore, 100 gram of \(CuSO_{4}\) will contain \(\frac{63.5\times 100g}{159.5}\) of Cu.
= \(\frac{63.5\times 100}{159.5}\)
=39.81 gram
Q8. The mass percent of iron and oxygen in an oxide of iron is 69.9 and 30.1 calculate the molecular formula of the oxide of iron. \(159.69\;g\;mol^{-1}\) is the given molar mass of an oxide.
Ans.
Here,
Mass percent of Fe = 69.9%
Mass percent of O = 30.1%
No. of moles of Fe present in oxide
= \(\frac{69.90}{55.85}\)
= 1.25
No. of moles of O present in oxide
= \(\frac{30.1}{16.0}\)
=1.88
Ratio of Fe to O in oxide,
= 1.25: 1.88
= \(\frac{1.25}{1.25}:\frac{1.88}{1.25}\)
=\(1:1.5\)
= \(2:3\)
Therefore, the empirical formula of oxide is \(Fe_{2}O_{3}\)
Empirical formula mass of \(Fe_{2}O_{3}\)
= [2(55.85) + 3(16.00)] gr
= 159.69 g
Therefore n = \(\frac{Molar\;mass}{Empirical\;formula\;mass}=\frac{159.69\;g}{159.7\;g}\)
= 0.999
= 1(approx)
The molecular formula of a compound can be obtained by multiplying n and the empirical formula.
Thus, the empirical of the given oxide is \(Fe_{2}O_{3}\) and n is 1.
Q9. Find out the atomic mass (average) of chlorine using the following data:
Percentage Natural Abundance | Molar Mass | |
\(_{}^{35}\textrm{Cl}\) | 75.77 | 34.9689 |
\(_{}^{37}\textrm{Cl}\) | 24.23 | 36.9659 |
Ans.
Average atomic mass of Cl.
=[(Fractional abundance of \(_{}^{35}\textrm{Cl}\))(molar mass of \(_{}^{35}\textrm{Cl}\))+(fractional abundance of \(_{}^{37}\textrm{Cl}\) )(Molar mass of \(_{}^{37}\textrm{Cl}\) )]
=[{(\(\frac{75.77}{100}(34.9689u)\) } + {(\(\frac{24.23}{100}(34.9659\;u)\) }]
= 26.4959 + 8.9568
= 35.4527 u
Therefore, the average atomic mass of Cl = 35.4527 u
Q10. In 3 moles of ethane (\(C_{2}H_{6}\)), calculate the given below:
(a) No. of moles of C- atoms
(b) No. of moles of H- atoms.
(c) No. of molecules of C_{2}H_{6}.
Ans.
(a) 1 mole \(C_{2}H_{6}\) contains two moles of C- atoms.
\(∴\) No. of moles of C- atoms in 3 moles of \(C_{2}H_{6}\).= 2 * 3
= 6
(b) 1 mole \(C_{2}H_{6}\) contains six moles of H- atoms.
\(∴\) No. of moles of C- atoms in 3 moles of \(C_{2}H_{6}\).= 3 * 6
= 18
(c) 1 mole \(C_{2}H_{6}\) contains six moles of H- atoms.
\(∴\) No. of molecules in 3 moles of \(C_{2}H_{6}\).= 3 * 6.023 * \(10^{23}\)
= 18.069 * \(10^{23}\)
Q11. What is the concentration of sugar \(C_{12}H_{22}O_{11}\) in mol \(L^{-1}\) if its 20 gram are dissolved in enough H_{2}O to make a final volume up to 2 Litre?
Ans.
Molarity (M) is as given by,
= \(\frac{Number\;of\;moles\;of\;solute}{Volume\;of\;solution\;in\;Litres}\)
= \(\frac{\frac{Mass\;of\;sugar}{Molar\;mass\;of\;sugar}}{2\;L}\)
= \(\frac{\frac{20\;g}{[(12\;\times \;12)\;+\;(1\;\times \;22)\;+\;(11\;\times \;16)]g]}}{2\;L}\)
= \(\frac{\frac{20\;g}{342\;g}}{2\;L}\)
= \(\frac{0.0585\;mol}{2\;L}\)
= 0.02925 mol\(L^{-1}\)
Therefore, Molar concentration = 0.02925 mol\(L^{-1}\)
Q12. The density of CH_{3}OH (methanol) is 0.793 kg \(L^{-1}\). For making 2.5 Litre of its 0.25 M solution what volume is needed?
Ans.)
Molar mass of \(CH_{3}OH\)
= (1 * 12) + (4 * 1) + (1 * 16)
= 32 g \(mol^{-1}\)
= 0.032 kg \(mol^{-1}\)
Molarity of the solution
= \(\frac{0.793\;kg\;L^{-1}}{0.032\;kg\;mol^{-1} }\)
= 24.78 mol\(L^{-1}\)
(From the definition of density)
\(M_{1}V_{1} = M_{2}V_{2}\) \(∴\) (24.78 mol\(L^{-1}\)) \(V_{1}\) = (2.5 L) (0.25 mol\(L^{-1}\)) \(V_{1}\) = 0.0252 Litre \(V_{1}\) = 25.22 Millilitre
Q13. Pressure is defined as force per unit area of the surface. Pascal, the SI unit of pressure is as given below:
1 Pa = 1 N \(m^{-2}\)
Assume that mass of air at the sea level is 1034 gcm^{-2}. Find out the pressure in Pascal.
Ans.
As per definition, pressure is force per unit area of the surface.
P = \(\frac{F}{A}\)
= \(\frac{1034\;g\;\times \;9.8\;ms^{-2}}{cm^{2}}\times \frac{1\;kg}{1000\;g}\times \frac{(100)^{2}\;cm^{2}}{1 m^{2}}\)
= 1.01332 × \(10^{5}\) kg \(m^{-1} s^{-2}\)
Now,
1 N = 1 kg m\(s^{-2}\)
Then,
1 Pa = 1 \(Nm^{-2}\)
= 1 \(kgm^{-2}\)\(s^{-2}\)
Pa = 1 \(kgm^{-1}\)\(s^{-2}\) \(∴\) Pressure (P) = 1.01332 × \(10^{5}\) Pa
Q14. Write SI unit for mass. Also define mass.
Ans.
Si Unit: Kilogram (kg)
Mass:
“The mass equal to the mass of the international prototype of kilogram is known as mass.”
Q15. Match the prefixes with their multiples in the table given below:
Prefixes | Multiples | |
(a) | femto | 10 |
(b) | giga | \(10^{-15}\) |
(c) | mega | \(10^{-6}\) |
(d) | deca | \(10^{9}\) |
(e) | micro | \(10^{6}\) |
Ans.
Prefixes | Multiples | |
(a) | femto | \(10^{-15}\) |
(b) | giga | \(10^{9}\) |
(c) | mega | \(10^{6}\) |
(d) | deca | 10 |
(e) | micro | \(10^{-6}\) |
Q16. What are significant figures?
Ans.
Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in experimented value.
e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3.
Therefore, “the total number of digits in a number with the Last digit the shows the uncertainty of the result is known as significant figures.”
Q17. A sample of drinking water was found to be highly contaminated with \(CHCl_{3}\), chloroform, which is carcinogenic. 15 ppm (by mass) was the level of contamination.
(a) Express in terms of percent by mass.
(b) Calculate the molality of chloroform in the given water sample.
Ans.
(a) 1 ppm = 1 part out of 1 million parts.
Mass percent of 15 ppm chloroform in H_{2}O
= \(\frac{15}{10^{6}} \times 100\)
= \(\approx\) 1.5 ×\(10^{-3}\) %
(b) 100 gram of the sample is having 1.5 ×\(10^{-3}\)g of \(CHCl_{3}\).
1000 gram of the sample is having 1.5 ×\(10^{-2}\)g of \(CHCl_{3}\).
\(∴\) Molality of \(CHCl_{3}\) in water= \(\frac{1.5 \;\times 10^{-2} \;g}{Molar \; mass \; of \; CHCl_{3}}\)
Molar mass (\(CHCl_{3}\))
= 12 + 1 + 3 (35.5)
= 119.5 gram \(mol^{-1}\)
Therefore, molality of \(CHCl_{3}\) I water
= 1.25 × \(10^{-4}\) m
Q18. Express the given number in scientific notation:
(a) 0.0047
(b) 235,000
(c)8009
(d)700.0
(e) 5.0013
Ans.
(a) 0.0047= 4.7 ×\(10^{-3}\)
(b) 235,000 = 2.35 ×\(10^{5}\)
(c) 8009= 8.009 ×\(10^{3}\)
(d) 700.0 = 7.000 ×\(10^{2}\)
(e) 5.0013 = 5.0013
Q19. Find the number of significant figures in the numbers given belowt.
(a) 0.0027
(b) 209
(c)6005
(d)136,000
(e) 900.0
(f)2.0035
Ans.
(i) 0.0027: 2 significant numbers.
(ii) 209: 3 significant numbers.
(iii)6005: 4 significant numbers.
(iv)136,000:3 significant numbers.
(v) 900.0: 4 significant numbers.
(vi)2.0035: 5 significant numbers.
Q20. Round up the given numbers upto 3 significant numbers.
(a) 35.217
(b) 11.4108
(c)0.05577
(d)2806
Ans.
(a) The number after round up is: 35.2
(b) The number after round up is: 11.4
(c)The number after round up is: 0.0560
(d)The number after round up is: 2810
Q21. When dioxygen and dinitrogen react together, they form various compounds. The information is given below:
Mass of dioxygen | Mass of dinitrogen | |
(i) | 16 g | 14 g |
(ii) | 32 g | 14 g |
(iii) | 32 g | 28 g |
(iv) | 80 g | 28 g |
(1) In the data given above, which chemical combination law is obeyed? Also give the statement of the law.
(2) Convert the following:
(a) 1 km =_____ mm = _____ pm
(b) 1 mg = _____ kg = _____ ng
(c) 1 mL = _____ L =_____ \(dm^{ 3 }\)
Ans.
(1) If we fix the mass of N_{2} at 28 g, the masses of N_{2} that will combine with the fixed mass of N_{2} are 32 gram, 64 gram, 32 gram and 80 gram.
The mass of O_{2} bear whole no. ratio of 1: 2: 2: 5. Therefore, the given information obeys the law of multiple proportions.
The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”
(2) Convert:
(a) 1 km = ____ mm = ____ pm
- 1 km = 1 km * \(\frac{ 1000 \; m }{ 1 \; km }\) × \(\frac{ 100 \; cm }{ 1 \; m }\) * \(\frac{ 10 \; mm }{ 1 \; cm }\)
- 1 km = 1 km * \(\frac{ 1000 \; m }{ 1 \; km }\) * \(\frac{1 \; pm}{10^{ -12 } \; m}\)
Therefore, 1 km = \(10^{ 6 }\) mm = \(10^{ 15 }\) pm
(b) 1 mg = ____ kg = ____ ng
- 1 mg = 1 mg * \(\frac{ 1 \; g }{ 1000 \; mg }\) * \(\frac{ 1 \; kg }{ 1000 \; g }\)
1 mg = \(10^{ -6 }\) kg
- 1 mg = 1 mg * \(\frac{ 1 \; g }{ 1000 \; mg }\) * \(\frac{ 1 \; ng }{ 10^{ -9 } \; g }\)
1 mg = \(10^{ 6 }\) ng
Therefore, 1 mg = \(10^{ -6 }\) kg = \(10^{ 6 }\) ng
(c) 1 mL = ____ L = ____ \(dm^{ 3 }\)
- 1 mL = 1 mL * \(\frac{1 \; L}{1000 \; mL}\)
1 mL = \(10^{ -3 }\) L
- 1 mL = 1 \(cm^{ 3 }\) = 1 * \(\frac{1 \; dm \; \times \; 1 \; dm \; \times \;1 \; dm }{10 \; cm \; \times \; 10 \; cm \; \times \; 10 \; cm } cm^{ 3 }\)
1 mL = \(10^{ -3 } dm^{ 3 }\)
Therefore, 1 mL = \(10^{ -3 }\)L = \(10^{ -3 }\) \(dm^{ 3 }\)
Q22. What is the distance covered by the light in 2 ns if the speed of light is 3 × \(10^{ 8 }\) \(ms^{ -1 }\)
Ans.
Time taken = 2 ns
= 2 × \(10^{ -9 }\) s
Now,
Speed of light = 3 × \(10^{ 8 }\) \(ms^{ -1 }\)
So,
Distance travelled in 2 ns = speed of light * time taken
= (3 × \(10^{ 8 }\))(2 × \(10^{ -9 }\))
= 6 × \(10^{ -1 }\) m
= 0.6 m
Q23. In the reaction given below:
\(X \; + \; Y_{ 2 } \; \rightarrow \; XY_{ 2 }\)
Find the limiting reagent if it is present in the reactions given below:
(a) 2 mol X + 3 mol Y
(b) 100 atoms of X + 100 molecules of Y
(c) 300 atoms of X + 200 molecules of Y
(d) 2.5 mol X + 5 mol Y
(e) 5 mol X + 2.5 mol Y
Ans.
Limiting reagent:
It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limiting the amt. of products formed.
(a) 2 mol X + 3 mol Y
1 mole of X reacts with 1 mole of Y. Similarly, 2 moles of X reacts with 2 moles of Y, so 1 mole of Y is unused. Hence, X is limiting agent.
(b) 100 atoms of X + 100 molecules of Y
1 atom of X reacts with 1 molecule of Y. Similarly, 100 atoms of X reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting agent.
(c) 300 atoms of X + 200 molecules of Y
1 atom of X reacts with 1 molecule of Y. Similarly, 200 atoms of X reacts with 200 molecules of Y, so 100 atoms of X are unused. Hence, Y is limiting agent.
(d) 2.5 mol X + 5 mol Y
1 mole of X reacts with 1 mole of Y. Similarly, 2.5 moles of X reacts with 2.5 moles of Y, so 2.5 mole of Y is unused. Hence, X is limiting agent.
(e) 5 mol X + 2.5 mol Y
1 mole of X reacts with 1 mole of Y. Similarly 2.5 moles of X reacts with 2 moles of Y, so 2.5 mole of X is unused. Hence, Y is limiting agent.
Q24. H_{2} and N_{2} react with each other to produce NH_{3} according to the given chemical equation
\(N_{ 2 }\; (g) \; + \; H_{ 2 }\; (g) \; \rightarrow \; 2NH_{ 3 }\;(g)\)
(a) What is the mass of \(NH_{ 3 }\) produced if \(2 \; \times \;10^{ 3 }\) g N_{2} reacts with \(1 \; \times \;10^{ 3 }\) g of H_{2}?
(b) Will the reactants N_{2} or H_{2 }remain unreacted?
(c) If any, then which one and give it’s mass.
Ans.
(a) Balance the given equation:
\(N_{ 2 }\;(g) \; + \; 3H_{ 2 } \;(g) \; \rightarrow \; 2NH_{ 3 }\;(g) \)
Thus, 1 mole (28 g) of N_{2} reacts with 3 mole (6 g) of H_{2} to give 2 mole (34 g) of \(NH_{ 3 }\).
\(2 \; \times \;10^{ 3 }\) g of N_{2} will react with \(\frac{ 6 g }{ 28 g } \; \times \; 2 \; \times \; 10^{ 3 }\) g H_{2}
\(2 \; \times \;10^{ 3 }\) g of N_{2} will react with 428.6 g of H_{2}.
Given:
Amt of H_{2} = \(1 \; \times \;10^{ 3 }\)
Therefore, \(N_{ 2 }\) is limiting reagent.
28 g of \(N_{ 2 }\) produces 34 g of \(NH_{ 3 }\)
Therefore, mass of \(NH_{ 3 }\) produced by 2000 g of \(N_{ 2 }\)
= \(\frac{ 34 \; g }{ 28 \; g } \; \times \; 2000\) g
(b) \(N_{ 2 }\) is limiting reagent and \(H_{ 2 }\) is the excess reagent. Therefore, \(H_{ 2 }\) will not react.
(c) Mass of H_{2} unreacted
= \(1 \; \times \;10^{ 3 }\) – 428.6 g
= 571.4 g
Q25. 0.50 mol \(Na_{ 2 }CO_{ 3 }\) and 0.50 M \(Na_{ 2 }CO_{ 3 }\) are different. How?
Ans.
Molar mass of \(Na_{ 2 }CO_{ 3 }\):
= (2 × 23) + 12 + (3 × 16)
= 106 g \(mol^{ -1 }\)
1 mole of \(Na_{ 2 }CO_{ 3 }\) means 106 g of \(Na_{ 2 }CO_{ 3 }\)
Therefore, 0.5 mol of \(Na_{ 2 }CO_{ 3 }\)
= \(\frac{ 106 \; g }{ 1 \; mol } \; \times \; 0.5 \; mol\) \(Na_{ 2 }CO_{ 3 }\)
= 53 g of \(Na_{ 2 }CO_{ 3 }\)
0.5 M of \(Na_{ 2 }CO_{ 3 }\) = 0.5 mol/L \(Na_{ 2 }CO_{ 3 }\)
Hence, 0.5 mol of \(Na_{ 2 }CO_{ 3 }\) is in 1 L of water or 53 g of \(Na_{ 2 }CO_{ 3 }\) is in 1 L of water.
Q26. If 10 volumes of dihydrogen gas react with 5 volumes of dioxygen gas, how many volumes of vapour would be obtained?
Ans.
Reaction:
\(2H_{ 2 }\;(g) \; + \; O_{ 2 }\; (g) \; \rightarrow \; 2H_{ 2 }O\; (g) \)
2 volumes of dihydrogen react with 1 volume of dioxygen to produce two volumes of vapour.
Hence, 10 volumes of dihydrogen will react with five volumes of dioxygen to produce 10 volumes of vapour.
Q27. Convert the given quantities into basic units:
(i) 29.7 pm
(ii) 16.15 pm
(iii) 25366 mg
Ans.
(i) 29.7 pm
1 pm = \(10^{ -12 } \; m\)
29.7 pm = 29.7 × \(10^{ -12 } \; m\)
= 2.97 × \(10^{ -11 } \; m\)
(ii) 16.15 pm
1 pm = \(10^{ -12 } \; m\)
16.15 pm = 16.15 × \(10^{ -12 } \; m\)
= 1.615 × \(10^{ -11 } \; m\)
(iii) 25366 mg
1 mg = \(10^{ -3 } \; g\)
25366 mg = 2.5366 × \(10^{ -1 } \) × \(10^{ -3 } \; kg\)
25366 mg = 2.5366 × \(10^{ -2 } \; kg\)
Q28. Which of the given below have the largest no. of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of \(Cl_{ 2 }\) (g)
Ans.
(i) 1 g Au (s)
= \(\frac{ 1 }{ 197 }\) mol of Au (s)
= \(\frac{ 6.022 \; \times \; 10^{ 23 } }{ 197 }\) atoms of Au (s)
= 3.06 \(\times \; 10^{ 21 }\) atoms of Au (s)
(ii) 1 g Na (s)
= \(\frac{ 1 }{ 23 }\) mol of Na (s)
= \(\frac{ 6.022 \; \times \; 10^{ 23 } }{ 23 }\) atoms of Na (s)
= 0.262 \(\times \; 10^{ 23 }\) atoms of Na (s)
= 26.2 \(\times \; 10^{ 21 }\) atoms of Na (s)
(iii) 1 g Li (s)
= \(\frac{ 1 }{ 7 }\) mol of Li (s)
= \(\frac{ 6.022 \; \times \; 10^{ 23 } }{ 7 }\) atoms of Li (s)
= 0.86 \(\times \; 10^{ 23 }\) atoms of Li (s)
= 86.0 \(\times \; 10^{ 21 }\) atoms of Li (s)
(iv)1 g of \(Cl_{ 2 }\) (g)
= \(\frac{ 1 }{ 71 }\) mol of \(Cl_{ 2 }\) (g)
(Molar mass of \(Cl_{ 2 }\) molecule = 35.5 × 2 = 71 g \(mol^{ -1 }\))
= \(\frac{ 6.022 \; \times \; 10^{ 23 } }{ 71 }\) atoms of \(Cl_{ 2 }\) (g)
= 0.0848 \(\times \; 10^{ 23 }\) atoms of \(Cl_{ 2 }\) (g)
= 8.48 \(\times \; 10^{ 21 }\) atoms of \(Cl_{ 2 }\) (g)
Therefore, 1 g of Li (s) will have the largest no. of atoms.
Q29. What is the molarity of the solution of ethanol in water in which the mole fraction of ethanol is 0.040?
(Assume the density of water to be 1)
Ans.
Mole fraction of \(C_{ 2 }H_{ 5 }OH\)
= \(\frac{Number \; of \; moles \; of \; C_{ 2 }H_{ 5 }OH}{Number \; of \; moles \; of \; solution}\)
0.040 = \(\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; n_{H_{ 2 }O}}\) ——(1)
No. of moles present in 1 L water:
\(n_{ H_{ 2 }O} \; = \; \frac{ 1000 \; g}{18 \; g \; mol^{ -1 }}\) \(n_{ H_{ 2 }O}\) = 55.55 mol
Substituting the value of \(n_{ H_{ 2 }O}\) in eq (1),
\(\frac{n_{C_{ 2 }H_{ 5 }OH}}{n_{C_{ 2 }H_{ 5 }OH} \; + \; 55.55}\) = 0.040 \(n_{C_{ 2 }H_{ 5 }OH}\) = 0.040\(n_{C_{ 2 }H_{ 5 }OH}\) + (0.040)(55.55)0.96\(n_{C_{ 2 }H_{ 5 }OH}\) = 2.222 mol
\(n_{C_{ 2 }H_{ 5 }OH}\) = \(\frac{ 2.222 }{ 0.96 } \; mol\) \(n_{C_{ 2 }H_{ 5 }OH}\) = 2.314 mol
Therefore, molarity of solution
= \(\frac{ 2.314 \; mol }{ 1 \; L }\)
= 2.314 M
Q30. Calculate the mass of 1 \(_{}^{ 12 }\textrm{ C }\) atom in g.
Ans.
1 mole of carbon atoms
= \(6.023 \; \times \; 10^{ 23 }\) atoms of carbon
= 12 g of carbon
Therefore, mass of 1 \(_{}^{ 12 }\textrm{ C }\) atom
= \(\frac{ 12 \; g }{ 6.022 \; \times \; 10^{ 23 }}\)
= \(1.993 \; \times \; 10^{ -23 } g\)
Q31. How many significant numbers should be present in answer of the given calculations?
(i) \(\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }\)
(ii) 5 × 5.365
(iii) 0.012 + 0.7864 + 0.0215
Ans.
(i) \(\frac{ 0.02856 \; \times \; 298.15 \; \times \; 0.112}{ 0.5785 }\)
Least precise no. of calculation = 0.112
Therefore, no. of significant numbers in the answer
= No. of significant numbers in the least precise no.
= 3
(ii) 5 × 5.365
Least precise no. of calculation = 5.365
Therefore, no. of significant numbers in the answer
= No. of significant numbers in 5.365
= 4
(iii) 0.012 + 0.7864 + 0.0215
As the least no. of decimal place in each term is 4, the no. of significant numbers in the answer is also 4.
Q32. Calculate molar mass of Argon isotopes according to the data given in the table.
Isotope | Molar mass | Abundance |
\(\, _{ 36 }\textrm{Ar}\) | 35.96755 \(g \; mol^{ -1 }\) | 0.337 % |
\(\, _{ 38 }\textrm{Ar}\) | 37.96272 \(g \; mol^{ -1 }\) | 0.063 % |
\(\, _{ 40 }\textrm{Ar}\) | 39.9624 \(g \; mol^{ -1 }\) | 99.600 % |
Ans.
Molar mass of Argon:
= [\(( 35.96755 \; \times \; \frac{ 0.337 }{ 100 })\) + \(( 37.96272 \; \times \; \frac{ 0.063 }{ 100 })\) + \(( 39.9624 \; \times \; \frac{ 99.600 }{ 100 })\)]
= [0.121 + 0.024 + 39.802] \(g \; mol^{ -1 }\)
= 39.947 \(g \; mol^{ -1 }\)
Q33. What is the number of atoms in the following compounds?
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He
Ans.
(i) 52 moles of Ar
1 mole of Ar = \(6.023 \; \times \; 10^{ 23 }\) atoms of Ar
Therefore, 52 mol of Ar = 52 × \(6.023 \; \times \; 10^{ 23 }\) atoms of Ar
= \(3.131 \; \times \; 10^{ 25 }\) atoms of Ar
(ii) 52 u of He
1 atom of He = 4 u of He
OR
4 u of He = 1 atom of He
1 u of He = \(\frac{ 1 }{ 4 }\) atom of He
52 u of He = \(\frac{ 52 }{ 4 }\) atom of He
= 13 atoms of He
(iii) 52 g of He
4 g of He = \(6.023 \; \times \; 10^{ 23 }\) atoms of He
52 g of He = \(\frac{ 6.023 \; \times \; 10^{ 23 } \; \times \;52 }{ 4 }\) atoms of He
= \(7.8286 \; \times \; 10^{ 24 }\) atoms of He
Q34. A welding fuel gas contains hydrogen and carbon. If we burn a small sample, we get 3.38 g of carbon dioxide and 0.69 g of water. A volume of 10 L (at STP) of this welding gas weighs 11.6 g.
Find:
(i) Empirical formula
(ii) Molar mass of the gas, and
(iii) Molecular formula
Ans.
(i) Empirical formula
1 mole of \(CO_{ 2 }\) contains 12 g of carbon
Therefore, 3.38 g of \(CO_{ 2 }\) will contain carbon
= \(\frac{ 12 \; g }{ 44 \; g } \; \times 3.38 \; g\)
= 0.9217 g
18 g of water contains 2 g of hydrogen
Therefore, 0.690 g of water will contain hydrogen
= \(\frac{ 2 \; g }{ 18 \; g } \; \times 0.690\)
= 0.0767 g
As hydrogen and carbon are the only elements of the compound. Now, the total mass is:
= 0.9217 g + 0.0767 g
= 0.9984 g
Therefore, % of C in the compound
= \(\frac{ 0.9217 \; g }{ 0.9984 \; g } \; \times 100\)
= 92.32 %
% of H in the compound
= \(\frac{ 0.0767 \; g }{ 0.9984 \; g } \; \times 100\)
= 7.68 %
Moles of C in the compound,
= \(\frac{ 92.32 }{ 12.00 }\)
= 7.69
Moles of H in the compound,
= \(\frac{ 7.68 }{ 1 }\)
= 7.68
Therefore, the ratio of carbon to hydrogen is,
7.69: 7.68
1: 1
Therefore, the empirical formula is CH.
(ii) Molar mass of the gas, and
Weight of 10 L of gas at STP = 11.6 g
Therefore, weight of 22.4 L of gas at STP
= \(\frac{ 11.6 \; g }{ 10 \; L } \; \times \; 22.4 \; L\)
= 25.984 g
\(\approx\) 26 g
(iii) Molecular formula
Empirical formula mass:
CH = 12 + 1
= 13 g
n = \(\frac{ Molar \; mass \; of \; gas}{Empirical \; formula \; mass \; of \; gas}\)
= \(\frac{ 26 \; g }{ 13 \; g}\)
= 2
Therefore, molecular formula is \((CH)_{ n }\) that is \(C_{ 2 }H_{ 2 }\).
Q35. Calcium carbonate reacts with aqueous HCl and gives \(CaCl_{ 2 }\) and \(CO_{ 2 }\) according to the reaction:
\(CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l) \)
Calculate the mass of \(CaCO_{ 3 }\) required to react completely with 25 mL of 0.75 M HCl?
Ans.
0.75 M of HCl
≡ 0.75 mol of HCl are present in 1 L of water
≡ [(0.75 mol) × (36.5 g mol–1 )] HCl is present in 1 L of water
≡ 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contins 27.375 g of HCl
Therefore, amt of HCl present in 25 mL of solution
= \(\frac{ 27.375 \; g }{ 1000 \; mL } \; \times \; 25 \; mL\)
= 0.6844 g
Given chemical reaction,
\(CaCO_{ 3 }\; (s) \; + \; 2 \; HCl\; (aq) \; \rightarrow \; CaCl_{ 2 }\;(aq) \; + \; CO_{ 2 }\; (g) \; + \; H_{ 2 }O\; (l) \)
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of \(CaCO_{ 3 }\) (100 g)
Therefore, amt of \(CaCO_{ 3 }\) that will react with 0.6844 g
= \(\frac{ 100 }{ 71 } \; \times \; 0.6844 \; g\)
= 0.9639 g
Q36. Chlorine is prepared by adding manganese dioxide with hydrochloric acid acc. to the reaction.
\(4 \; HCl\;(aq) \; + \; MnO_{ 2 }\;(s) \; \rightarrow \; 2 \; H_{ 2 }O\; (l) \; + \; MnCl_{ 2 }\; (aq) \; + \; Cl_{ 2 }\; (g) \)
How many grams of HCl react with 5 g of manganese dioxide?
Ans.
1 mol of \(MnO_{2}\) = 55 + 2 × 16 = 87 g
4 mol of HCl = 4 × 36.5 = 146 g
1 mol of \(MnO_{2}\) reacts with 4 mol of HCl
5 g of \(MnO_{ 2 }\)will react with:
= \(\frac{146 \; g}{87 \; g} \; \times \; 5 \; g\) HCl
= 8.4 g HCl
Therefore, 8.4 g of HCl will react with 5 g of \(MnO_{2}\).
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