5.5 g of a mixture of FeSO4.7H2O and Fe2(SO4)3 9H2O requires 5.4 mL of 0.1N KMnO4 solution for complete oxidation. Calculate the number of moles of Fe2(SO4)39H2O in the mixture. A. 0.0095 B. 0.15 C. 0.0952 D. 1.52 Solution: (A) Milliequivalent of KMnO4 = 5.4 x 0.1 = 0.54 milliequivalent = 0.54... View Article
In the analysis of a 0.0500 g sample of feldspar, a mixture of the chlorides of sodium and potassium is obtained, which weighs 0.1180 g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451 g of silver chloride. What is the percentage of sodium oxide and potassium oxide in feldspar? (a) 10.62% Na2O, 3.58% K2O (b) 3.58% Na2O, 10.62% K2O (c) 10.62% Na2O, 35.8% K2O (d) 35.8% Na2O, 10.62% K2O Solution: (B) Suppose the... View Article
Some statements are given below based on the pictures. Identify true and false statements. (i) ‘P’ and ‘Q’ both indicate precision and accuracy. (ii) ‘Q’ indicates precision and accuracy while... View Article
Calculate the number of sulphate ions in 100 mL of 0.001 M ammonium sulphate solution. A. 6.022 x 10-19 B. 6.022 x 1019 C. 6.022 x 1020 D. 6.022 x 10-20 Solution: (B) No of moles of (NH4)2 SO4 = molarity x vol(L) = 0.001... View Article
Calculate the concentration of nitric acid in moles per litre in a sample that has a density of 1.41 g mL-1 and the mass per cent of nitric acid in it being 69%. A. 15.44 M B. 0.064 M C. 0.077 M D. 12.87 M Solution: (A) The mass per cent of 69% means that 100g of nitric acid solution containS 69... View Article
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass. A. FeO B. Fe2O3 C. Fe3O4 D. Fe3O2 Solution: (B) The empirical formula of a compound may be defined as the formula which gives the... View Article
Calculate the mass per cent of Na and S in sodium sulphate. A. Na = 16.2%, S = 22.54% B. Na = 32.39%, S = 11.26% C. Na = 22.54%, S = 32.39% D. Na = 32.39%, S = 22.54% Solution: (A) Molar mass of... View Article
A naturally occurring Boron consists of two isotopes having atomic masses 10.01 and 11.01 respectively. Calculate the percentage of both the isotopes in natural Boron (Atomic mass of natural Boron = 10.81) A. 20% and 80% B. 20% and 80% C. 25% and 75% D. 75% and 25% Solution: (A) Let x% be the percentage of an isotope with atomic weights... View Article
4.88g of KClO3 when heated produced 1.92g of O2 and 2.96g of KCl. Which of the following statements regarding the experiment is correct? A. Law of conservation of mass B. Law of multiple proportions C. Law of definite proportions D. Avogadro’s Law Solution: (A)... View Article
Identify the incorrect unit conversion factor. A. 1 cm3 / 1 m L B. 1 cm / 10 mm C. 60 s / 1 min D. None of these Solution: (D) None of the above. All are correct options.... View Article
Identify and pick the incorrect statement. A. The constituents of a compound cannot be separated into simpler substances by physical methods. B. An element consists of only one type of... View Article
How many moles of electron weight one kilogram? A. 6.023 × 1023 B. 1 / 9.108 × 1023 C. 6.023 / 9.108 × 1054 D. 1 / 9.108 × 6.023 × 108 Solution: (D) Mass... View Article
In the standardisation of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is; A. (molecular weight) / 2 B. (molecular weight) / 6 C. (molecular weight) / 3 D. same as molecular weight Solution: (B) Equivalent... View Article
In an organic compound of molar mass 108 g mol-1 C, H and N atoms are present in 9 : 1 : 3.5 by weight. The molecular formula can be; A. C6H8N2 B. C7H10N C. C5H6N3 D. C4H18N3 Solution: (A) Given, Molar mass = 108 g/mol. Therefore, The mass of C present in one... View Article
A gaseous mixture contains CH4 and C2H6 in equimolecular proportion. The weight of 2.24 litres of this mixture at NTP is: A. 3 g B. 4.6 g C. 1.6 g D. 2.3 g Solution: (D) Mol wt of CH4 = 16 g/mol Mol wt of C2H6 = 30 g/mol Total = 46 g/mol ⇒1 Mol... View Article
An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is: A. 40 mL B. 20 mL C. 10 mL D. 4 mL Solution: (A) n-factor for H2C2O4.2H2O = 2 Normality = weight × 2 × 1000 / molecular weight... View Article
100 mL of 0.1N hypo decolourized iodine by the addition of x g of crystal copper sulphate to an excess of KI. The value of x is: A. 5.0 g B. 1.25 g C. 2.5 g D. 4 g Solution: (C) Let us first find the valence factor and write down the chemical reaction: 2CuSO4 +... View Article
n g of substance X reacts with m g of substance Y to form p g of substance R and q g of substance S. This reaction can be represented as X + Y = R + S. The relation which can be established in the amounts of the reactants and the products will be: A. n − m = p − q B. n + m = p + q C. n = m D. p = q Solution: (B) X + Y ⇌ R + S ng mg pg qg N + m = p + q considering... View Article
If the density of water is 1 g cm−3 then the volume occupied by one molecule of water is approximately; A. 18 cm3 B. 22400 cm3 C. 6.02 × 10−23 cm3 D. 3.0 × 10−23 cm3 Solution: (D) We know, Density = mass / volume... View Article
3.92 g of ferrous ammonium sulphate crystal are dissolved in 100 mL of water, 20 mL of this solution requires 18mL of potassium permanganate during titration for complete oxidation. The weight of KMnO4 present in one litre of the solution is: A. 34.76 g B. 12.38 g C. 1.238 g D. 3.476 g Solution: (D) Molecular weight of Ferrous ammonium sulphate ((NH4)2Fe(SO4)2.6H2O) = 392... View Article
