NCERT Solutions for Class 12 Chemistry Chapter 2 Solutions provides accurate and adequate chemistry knowledge. Chemistry class 12 solutions chapter 2 pdf can be utilised by the students to prepare for the examination and to solve the questions of class 12 chemistry chapter 2 exercise solutions along with exemplar problems, MCQ’S, short and long answer questions. These solutions can help the students to clear all their doubts related to this chapter very easily.
A solution is a homogeneous mixture of two or more than two components. Solved question and answer on the chapter solutions are given by our subject experts. These solutions will help the students to understand types of solution, expressing the concentration of the solution, solubility, ideal and nonideal solutions, colligative properties and abnormal molar mass. Chemistry class 12 questions and solutions chapter 2 given here are very simple and easy to understand.
Class 12 NCERT Solutions for Solutions
This chapter holds approximately 5 marks in the board examination. This chapter 2 “solution” teaches about the types of solutions, the concentration of solutions, solubility of solids and gases in a liquid, the vapour pressure of liquid solutions, Raoult’s law, ideal and nonideal solutions, colligative properties and determination of molar masses.
Subtopics for Class 12 Chemistry Chapter 2 – Solutions
 Types of Solutions
 Expressing Concentration of Solutions

Solubility
 The solubility of a Solid in a Liquid
 The solubility of a Gas in a Liquid

Vapour Pressure of Liquid Solutions
 Vapour Pressure of LiquidLiquid Solutions
 Raoult’s Law as a special case of Henry’s Law
 Vapour Pressure of Solutions of Solids in Liquids

Ideal and Nonideal Solutions
 Ideal Solutions
 Nonideal Solutions

Colligative Properties and Determination of Molar Mass
 Relative Lowering of Vapour Pressure
 Elevation of Boiling Point
 Depression of Freezing Point
 Osmosis and Osmotic Pressure
 Abnormal Molar Masses
NCERT chemistry book includes a chapter on solutions to introduce various important concepts to the students. In this students learn to determine the molarity, molality and mole fraction of solutions, know about Henry’s law constant, mass percentage, etc. The topics provided in the NCERT books are not only important for the class 12 board examination but also for the competitive exams like JEE Mains and NEET. JEE Mains is a national level engineering entrance examination and NEET is a national level examination to take admission in best medical colleges of India.
Solving the NCERT questions will help you to understand the topics in a better and simple way. Sometimes the questions are also asked in the JEE Mains and NEET examinations.
Class 12 Chemistry NCERT Solutions (Solutions) – Important Questions
Q 2.1) If 22 g of benzene is dissolved in 122 g of carbon tetrachloride, determine the mass percentage of carbon tetrachloride (CCl_{4}) and benzene (C_{6}H_{6}).
Answer 2.1:
Mass percentage of Benzene (C_{6}H_{6}) =
=
=
= 15.28%
Mass percentage of Carbon Tetrachloride (CCl_{4}) =
=
=
= 84.72%
Q 2.2) If benzene in solution containing 30% by mass in carbon tetrachloride, calculate the mole fraction of benzene?
Answer 2.2:
Assume the mass of benzene be 30 g in the total mass of the solution of 100 g.
Mass of CCl_{4} = (100 − 30) g
= 70 g
Molar mass of benzene (C_{6}H_{6}) = (6 × 12 + 6 × 1) g
= 78 g
Therefore, Number of moles of C_{6}H_{6} =
= 0.3846 mol
Molar mass of CCl_{4} = 1 x 12 + 4 x 355 = 154 g
Therefore, Number of moles of CCl_{4} =
= 0.4545 mol
Thus, the mole fraction of C_{6}H_{6} is given as:
=
= 0.458
Q 2.3) Determine the molarity of each of the solutions given below:
(a) 30 g of Co(NO)_{3}. 6H_{2}O in 4.3 L of solution
(b) 30 mL of 0.5 M H_{2}SO_{4} diluted to 500 mL.
Answer 2.3:
We know that,
Molarity =
(a) Molar mass of Co(NO)_{3}. 6H_{2}O = 59 + 2 (14 + 3 x 16) + 6 x 18 = 291 g
Therefore, Moles of Co(NO)_{3}. 6H_{2}O =
= 0.103 mol
Therefore, molarity =
= 0.023 M
(b) Number of moles present in 1000 mL of 0.5 M H_{2}SO_{4} = 0.5 mol
Therefore, Number of moles present in 30 mL of 0.5 M H_{2}SO_{4} =
= 0.015 mol
Therefore, molarity =
= 0.03 M
Q 2.4) To make 2.5 kg of 0.25 molar aqueous solution, determine the mass of urea (NH_{2}CONH_{2}) that is required.
Answer 2.4:
Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g
0.25 molar aqueous solution of urea means:
1000 g of water contains 0.25 mol = (0.25 × 60) g of urea = 15 g of urea
That is,
(1000 + 15) g of solution contains 15 g of urea
Therefore, 2.5 kg (2500 g) of solution contains =
= 36.95 g
= 37 g of urea (approx.)
Hence, mass of Urea required is 37 g.
Q 2.5) If 1.202 g
(a) Molality of KI
(b) Molarity of KI
(c) Mole fraction of KI
Answer 2.5:
(a) Molar mass of KI = 39 + 127 = 166 g
20% aqueous solution of KI means 20 g of KI is present in 100 g of solution.
That is,
20 g of KI is present in (100 – 20) g of water = 80 g of water
Therefore, molality of the solution =
=
= 1.506 m
= 1.51 m (approx.)
(b) It is given that the destiny of the solution = 1.202
Volume of 100 g solution =
=
= 83.19 mL
=
Therefore, molarity of the solution =
= 1.45 M
(c) Moles of KI =
Moles of water =
Therefore, mole =
Fraction of KI =
= 0.0263
Q 2.6) Calculate Henry’s law constant when the solubility of H_{2}S( a toxic gas with rotten egg like smell) in water at STP is 0.195 m
Answer 2.6:
It is given that the solubility of H_{2}S in water at STP is 0.195 m, i.e., 0.195 mol of H_{2}S is dissolve in 1000 g of water.
Moles of water =
= 55.56 mol
Therefore, Mole fraction of H_{2}S, x =
=
= 0.0035
At STP, pressure (p) = 0.987 bar
According to Henry’s law: p =
=>
=
= 282 bar
Q 2.7) A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Answer 2.7:
Total amount of solute present in the mixture is given by,
= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage of the solute in the resulting solution =
= 33.57%
And, mass percentage of the solvent in the resulting solution is:
= (100 – 33.57) %
= 66.43%
Q 2.8) The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.
Answer 2.8:
It is given that:
According to Raoult’s law:
Therefore, total pressure,
=>
=>
=>
=> 600 = (450 – 700) x_{A} + 700
=> –100 = –250x_{A}
=> x_{A} = 0.4
Therefore,
Now,
= 450 x 0.4 = 180 mm of Hg
= 700 x 0.6 = 420 mm of Hg
Now, in the vapour phase: Mole fraction of liquid A =
=
=
= 0.30
And, mole fraction of liquid B = 1 – 0.30 = 0.70
Q 2.9) Find the vapor pressure of water and its relative lowering in the solution which is 50 g of urea (NH_{2}CONH_{2}) dissolved in 850 g of water. (Vapor pressure of pure water at 298 K is 23.8 mm Hg)
Answer 2.9:
It is given that vapour pressure of water,
Weight of water taken,
Weight of urea taken,
Molecular weight of water,
Molecular weight of urea,
Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p_{1}.
Now, from Raoult’s law, we have:
=>
=>
=>
=>
=>
Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.
Q 2.10) How much of sucrose is to be added to 500 g of water such that it boils at 100°C if the molar elevation constant for water is 0.52 K kg mol^{1} and the boiling point of water at 750 mm Hg is 99.63°C.
Answer 2.10:
Here, elevation of boiling point
= 0.37 K
Mass of water,
Molar mass of sucrose (C_{12}H_{22}O_{11}),
= 342 g
Molar elevation constant, K_{b} = 0.52 K kg
We know that:
=>
=
= 121.67 g (approximately)
Hence, the amount of sucrose that is to be added is 121.67 g
Q 2.11) To lower the melting point of 75 g of acetic acid by 1.5^{0}C, how much mass of ascorbic acid is needed to be dissolved in the solution where K_{t} = 3.9 K kg
Answer 2.11:
Mass of acetic acid (w_{1}) = 75 g
Molar mass of ascorbic acid (C_{6}H_{8}O_{6}), M_{2} = 6 x 12 + 8 x 1 + 6 x 16 = 176 g
Lowering the melting point
We know that:
=>
=
= 5.08 g (approx)
Hence, the amount of ascorbic acid needed to be dissolved is 5.08 g.
Q 2.12) If a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C, calculate the osmotic pressure in Pascal exerted by it?
Answer 2.12:
It is given that:
Volume of water (V) = 450 mL = 0.45 L
Temperature (T) = 37 + 273 = 310 K
Number of moles of the polymer, n =
We know that:
Osmotic pressure,
=
= 30.98 Pa
= 31 Pa (approx)
Q 2.13) The partial pressure of ethane over a solution containing 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas?
Answer 2.13:
Molar mass of ethane (C_{2}H_{6}) = 2 x 12 + 6 x 1 = 30 g
Therefore, number of moles present in
=
Let ‘x’ be the number of moles of the solvent, According to Henry’s law,
=> 1 bar =
=> 1 bar =
=>
Number of moles present in
=
According to Henry’s law,
=
=
= 0.764 bar
Hence, partial pressure of the gas shall be 0.764 bar.
Q 2.14) What is meant by positive and negative deviations from Raoult’s law and how is the sign of
Answer 2.14:
According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (nonideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.
Vapour pressure of a twocomponent solution showing positive deviation from Raoult’s law
Vapour pressure of a twocomponent solution showing negative deviation from Raoult’s law
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
In the case of solutions showing positive deviations, absorption of heat takes place.
In the case of solutions showing negative deviations, evolution of heat takes place.
Q 2.15) An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Answer 2.15:
Vapour pressure of the solution at normal boiling point,
Vapour pressure of pure water at normal boiling point,
Mass of solute, w_{2} = 2 g
Mass of solvent (water), M_{1} = 18 g
According to Raoult’s law,
=>
=>
=>
= 41.35 g
Hence, 41.35 g
Q 2.16) Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Answer 2.16:
Vapour pressure of heptanes,
Vapour pressure of octane,
We know that,
Molar mass of heptanes (C_{7}H_{16}) = 7 x 12 + 16 x 1 = 100 g
Therefore, number of moles of heptane =
Molar mass of octane (C_{8}H_{18}) = 8 x 12 + 18 x 1 = 114 g
Therefore, number of moles of octane =
Mole fraction of heptane,
And, mole fraction of octane,
Now, partial pressure of heptane,
= 0.456 x 105.2
= 47.97 KPa
Partial pressure of octane,
= 0.544 x 46.8
= 25.46 KPa
Hence, vapour pressure of solution,
= 47.97 + 25.46
= 73.43 KPa
Q 2.17) The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a nonvolatile solute in it.
Answer 2.17:
1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water).
Molar mass of water = 18 g
Therefore, number of moles present in 1000 g of water =
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
It is given that,
Vapour pressure of water,
Applying the relation,
=>
=> 12.3 – p_{1} = 0.2177
=> p_{1} = 12.0823
= 12.08 KPa (approx)
Hence, the vapour pressure of the solution is 12.08 KPa.
Q 2.18) Calculate the mass of a nonvolatile solute (molar mass 40 g mol–1) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Answer 2.18:
Let
Then, after dissolving the nonvolatile solute the vapour pressure of octane is
Molar mass of solute, M_{2} = 40 g
Mass of octane, w_{1} = 114 g
Molar mass of octane, (C_{8}H_{18}), M_{1} = 8 x 12 + 18 x 1 = 114 g
Applying the relation,
=>
=>
=> 0.2 =
=> w_{2} = 8 g
Hence, the required mass of the solute is 8 g.
Q 2.19) A solution containing 30 g of nonvolatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: (i) molar mass of the solute (ii) vapour pressure of water at 298 K.
Answer 2.19:
(i) Let, the molar mass of the solute be M g
Now, the number of moles of solvent (water),
And, the number of moles of solute,
p_{1} = 2.8 KPa
Applying the relation:
=>
=>
=>
=>
=>
=>
=>
After the addition of 18 g of water:
Again applying the relation:
=>
=>
=>
=>
=>
=>
=>
Dividing equation (i) by (ii), we have:
=>
=>
=> 87M + 435 = 84M + 504
=> 3M = 69
=> M = 23 g
Therefore, 23 g
(ii) Putting the value of ‘M’ in equation (i), we have:
=>
=>
Hence, 3.53 KPa is the vapour pressure of water at 298 K.
Q 2.20) A 5% solution (by mass) of cane sugar in water has freezing point of 271K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K
Answer 2.20:
Molar mass of sugar (C_{12}H_{22}O_{11}) = 12 x 12 + 22 x 1 + 11 x 16 = 342 g
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g
= 95 g of water.
Now, number of moles of cane sugar =
Therefore, molality of the solution,
Applying the relation,
=>
=
= 13.99 K kg
Molar mass of glucose (C_{6}H_{12}O_{6}) = 6 x 12 + 12 x 1 + 6 x 16 = 180 g
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
Therefore, number of moles of glucose =
Therefore, molality of the solution, m =
= 0.2926 mol
Applying the relation:
=
= 4.09 K (approx)
Hence, the freezing point of 5 % glucose solution is (273.15 – 4.09) K = 269.06 K.
Q 2.21) Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB4 lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol–1. Calculate atomic masses of A and B.
Answer 2.21:
We know that,
Then,
= 110.87 g
= 196.15 g
Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
x + 2y = 110.87 …(i)
x + 4y = 196.15 …(ii)
Subtracting equation (i) from (ii), we have
2y = 85.28
=> y = 42.64
Putting the value of ‘y’ in equation (1), we have:
x + 2 (42.64) = 110.87
=> x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
Q 2.22) At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer 2.22:
Given:
T = 300 K
n = 1.52 bar
R = 0.083 bar L
Applying the relation, n = CRT
=> C =
=
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.
Q 2.23) Suggest the most important type of intermolecular attractive interaction in
the following pairs.
(i) nhexane and noctane
(ii) I2 and CCl4
(iii) NaClO4 and water
(iv) methanol and acetone
(v) acetonitrile (CH3CN) and acetone (C3H6O).
Answer 2.23:
(i) Van der Wall’s forces of attraction.
(ii) Van der Wall’s forces of attraction.
(iii) Iondiople interaction.
(iv) Dipoledipole interaction.
(v) Dipoledipole interaction.
Q 2.24) Based on solutesolvent interactions, arrange the following in order of increasing solubility in noctane and explain. Cyclohexane, KCl, CH3OH, CH3CN.
Answer 2.24:
noctane is a nonpolar solvent. Therefore, the solubility of a nonpolar solute is more than that of a polar solute in the noctane.
The order of increasing polarity is:
Cyclohexane < CH_{3}CN < CH_{3}OH < KCl
Therefore, the order of increasing solubility is:
KCl < CH_{3}OH < CH_{3}CN < Cyclohexane
Q 2.25) Amongst the following compounds, identify which are insoluble, partially
soluble and highly soluble in water?
(i) phenol (ii) toluene (iii) formic acid
(iv) ethylene glycol (v) chloroform (vi) pentanol.
Answer 2.25:
(i) Phenol (C_{6}H_{5}OH) has the polar group −OH and nonpolar group –C_{6}H_{5}. Thus, phenol is partially soluble in water.
(ii) Toluene (C_{6}H_{5}−CH_{3}) has no polar groups. Thus, toluene is insoluble in water.
(iii) Formic acid (HCOOH) has the polar group −OH and can form Hbond with water.
Thus, formic acid is highly soluble in water.
(iv)Ethylene glycol has polar −OH group and can form H−bond. Thus, it is highly soluble in water.
(v) Chloroform is insoluble in water.
(vi)Pentanol (C_{5}H_{11}OH) has polar −OH group, but it also contains a very bulky nonpolar −C5H11 group. Thus, pentanol is partially soluble in water.
Q 2.26) If the density of some lake water is 1.25g mL–1 and contains 92 g of Na+ ions per kg of water, calculate the molarity of Na+ ions in the lake
Answer 2.26:
Number of moles present in 92 g of Na^{+} ions =
= 4 mol
Therefore, molality of Na^{+} ions in the lake =
= 4 m
Q 2.27) If the solubility product of CuS is 6 × 10–16, calculate the maximum molarity of
CuS in aqueous solution.
Answer 2.27:
Solubility product of CuS,
Let s be the solubility of CuS in mol L^{1}.
Now, s s
= s x s
= s^{2}
Then, we have,
=>
=
Hence,
Q 2.28) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when
6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
Answer 2.28:
6.5 g of C_{9}H_{8}O_{4} is dissolved in 450 g of CH_{3}CN.
Then, total mass of the solution = (6.5 + 450) g = 456.5 g
Therefore, mass percentage of C_{9}H_{8}O_{4} =
= 1.424%
Q 2.29) Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10–3 m aqueous solution required for the above dose.
Answer 2.29:
The molar mass of nalorphene (C_{19}H_{21}NO_{3}) = 19 x 12 + 21 x 1 + 1 x 14 + 3 x 16 = 311 g mol^{1}
In
1 kg (1000 g) of water contains
= 0.4665 g
Therefore, total mass of the solution = (1000 + 0.4665) g = 1000.4665 g
This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.
Therefore, the mass of the solution containing 1.5 mg of nalorphene is:
= 3.22 g
Hence, 3.22 g is the required mass of aqueous solution.
Q 2.30) Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250
mL of 0.15 M solution in methanol.
Answer 2.30:
0.15 M solution of benzoic acid in methanol means,
1000 mL of solution contains 0.15 mol of benzoic acid
Therefore, 250 mL of solution contains
= 0.0375 mol of benzoic acid
Molar mass of benzoic acid (C_{6}H_{5}COOH) = 7 x 12 + 6 x 1 + 2 x 16 = 122 g mol^{1}
Hence, required benzoic acid = 0.0375 mol x 122 g mol^{1} = 4.575 g
Q 2.31) The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer 2.31:
Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H^{+} ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:
Acetic acid < trichloroacetic acid < trifluoroacetic acid
Q 2.32) Calculate the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Kf = 1.86 K kg mol–1
Answer 2.32:
Molar mass of CH_{3}CH_{2}CHCICOOH = 15 + 14 + 13 + 35.5 + 12 + 16 + 16 + 1
= 122.5 g mol^{1}
Therefore, No. of moles present in 10 g of CH_{3}CH_{2}CHCICOOH =
= 0.0816 mol
It is given that 10 g of CH_{3}CH_{2}CHCICOOH is added to 250 g of water.
Therefore, Molality of the solution, CH_{3}CH_{2}CHCICOOH =
= 0.3264 mol kg^{1}
Let ’a’ be the degree of dissociation of CH_{3}CH_{2}CHCICOOH.
CH_{3}CH_{2}CHCICOOH undergoes dissociation according to the following equation:
=
Since a is very small with respect to 1,
Now,
=>
=>
=
= 0.0655
Again,
Total moles of equilibrium = 1 – a + a + a = 1 + a
=
= 1 + 0.0655
= 1.0655
Hence, the depression in the freezing point of water is given as:
=
= 0.65 K
Q 2.33) 19.5 g of CH2 FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.00 C. Calculate the van’t Hoff factor and dissociation constant of fluoroacetic acid
Answer 2.33:
Given:
w_{1} = 500 g
w_{2} = 19.5 g
K_{f} = 1.86 K kg
We know that:
=
=
Therefore, observed molar mass of CH_{2}FCOOH,
The calculated molar mass of CH_{2}FCOOH,
Therefore, van’t Hoff factor,
=
= 1.0753
Let ‘a’ be the degree of dissociation of CH_{2}FCOOH
=> i = 1 +
=>
= 1.0753 – 1
= 0.0753
Now, the value of K_{a} is given as:
=
=
Taking the volume of the solution as 500 mL, we have the concentration:
C =
= 0.5 M
Therefore,
=
=
= 0.00307 (approx)
=
Q 2.34) Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer 2.34:
Vapour pressure of water,
Mass of glucose, w_{2} = 25 g
Mass of water, w_{1} = 450 g
We know that,
Molar mass of glucose (C_{6}H_{12}O_{6}), M_{2} = 6 x 12 + 12 x 1 + 6 x 16 = 180 g
Molar mass of water, M_{1} = 18 g
Then, number of moles of glucose,
= 0.139 mol
And, number of moles of water,
= 25 mol
We know that,
=>