1. The empirical formula of an organic compound containing carbon and hydrogen is \( CH_{2} \). The mass of one litre of this organic gas is exactly equal to that of one litre of \( N_{2} \). Therefore, the molecular formula of the organic gas is:

• (a) \( C_{2}H_{4} \)
• (b) \( C_{3}H_{6} \)
• (c) \( C_{8}H_{12} \)
• (d) \( C_{4}H_{8} \)

Correct Answer: (a)

Explanation: According to Avogadro’s hypothesis, the mass of 1 L of gas equals the mass of 1 L of \( N_{2} \), meaning their molecular masses are equal. The molecular mass of \( N_{2} \) is 28. The molecular mass of the gas is therefore 28, which corresponds to the formula \( C_{2}H_{4} \) (\( 2 \times 12 + 4 \times 1 = 28 \)).

2. Match the following:

• (a) A-(iv), B-(i), (C)-(ii), (D)-(v), (E)-(iii)
• (b) A-(ii), B-(i), (C)-(iv), (D)-(iii), (E)-(v)
• (c) A-(i), B-(ii), (C)-(iii), (D)-(iv), (E)-(v)
• (d) A-(v), B-(iii), (C)-(iv), (D)-(ii), (E)-(i)

Correct Answer: (a)

Explanation: This option correctly matches the physical quantities to their standard units and conversion factors.

3. The relative abundance of two isotopes of atomic weight 85 and 87 is 75% and 25% respectively. The average atomic weight of element is:

• (a) 86.0
• (b) 40.0
• (c) 85.5
• (d) 75.5

Correct Answer: (c)

Explanation: Average atomic weight is calculated as: \[ \frac{R.A.(1) \times \text{At. mass}(1) + R.A.(2) \times \text{At. mass}(2)}{R.A.(1) + R.A.(2)} \] \[ = \frac{75 \times 85 + 25 \times 87}{100} = \frac{6375 + 2175}{100} = 85.50 \].

4. Two students X and Y report the weight of the same substance as 4.0 g and 4.00 g respectively. Which of the following statement is correct?

• (a) Both are equally accurate
• (b) X is more accurate than Y
• (c) Y is more accurate than X
• (d) Both are inaccurate scientifically

Correct Answer: (c)

Explanation: 4.00 is more accurate than 4.0 because the former has three significant figures while the latter only has two.

5. 112.0 mL of \( NO_{2} \) at STP was liquefied, the density of the liquid being 1.15 g \( mL^{-1} \). Calculate the volume and the number of molecules in the liquid \( NO_{2} \).

• (a) 0.10 mL and \( 3.01 \times 10^{22} \)
• (b) 0.20 mL and \( 3.01 \times 10^{21} \)
• (c) 0.20 mL and \( 6.02 \times 10^{23} \)
• (d) 0.10 mL and \( 6.02 \times 10^{21} \)

Correct Answer: (b)

Explanation: At STP, 22400 mL of \( NO_{2} \equiv 46 \) g. Therefore, 112.0 mL = \( \frac{112.0 \times 46.0}{22400} = 0.23 \) g. Volume \( V_{NO_{2}} = \frac{\text{mass}}{\text{density}} = \frac{0.23}{1.15} = 0.20 \) mL. Number of molecules = \( \frac{0.23}{46} \times 6.02 \times 10^{23} = 3.01 \times 10^{21} \).

6. The specific heat of a metal is 0.16. Its approximate atomic weight would be:

• (a) 32
• (b) 16
• (c) 40
• (d) 64

Correct Answer: (c)

Explanation: Using Dulong and Petit’s Law: Approximate atomic weight = 6.4 / specific heat = \( 6.4 / 0.16 = 40 \).

7. 14 g of element X combine with 16 g of oxygen. On the basis of this information, which of the following is a correct statement? (Atomic weight of oxygen = 16)

• (a) The element of X could have an atomic weight of 7 and its oxide the formula XO.
• (b) The element X could have an atomic weight of 14 and its oxide the formula \( X_{2}O \).
• (c) The element X could have an atomic weight of 7 and its oxide the formula \( X_{2}O \).
• (d) The element X could have the atomic weight of 14 and its oxide the formula \( XO_{2} \).

Correct Answer: (c)

Explanation: If the atomic weight of X is 7, then 14 g of X is 2 moles. 16 g of oxygen is 1 mole. Thus, 2 moles of X combine with 1 mole of O, resulting in the molecular oxide \( X_{2}O \).

8. The maximum number of molecules are present in:

• (a) 15 L of \( H_{2} \) gas at S.T.P.
• (b) 5 L of \( N_{2} \) gas at S.T.P.
• (c) 0.5 g of \( H_{2} \) gas
• (d) 10 g of \( O_{2} \) gas

Correct Answer: (a)

Explanation: At STP, 22.4 L of gas contains \( 6.02 \times 10^{23} \) molecules. Calculations show: (a) \( 15 L H_{2} = 4.03 \times 10^{23} \) molecules; (b) \( 5 L N_{2} = 1.344 \times 10^{23} \); (c) \( 0.5 g H_{2} = 1.505 \times 10^{23} \); (d) \( 10 g O_{2} = 1.88 \times 10^{23} \). Option (a) contains the most molecules.

9. A gas is found to have the formula \( (CO)_{x} \). Its vapour density is 70. The value of x must be:

• (a) 7
• (b) 4
• (c) 5
• (d) 6

Correct Answer: (c)

Explanation: Molecular mass = \( 2 \times \text{vapour density} = 2 \times 70 = 140 \). The empirical formula (E.F.) mass for CO is \( 12 + 16 = 28 \). Therefore, \( x = \frac{140}{28} = 5 \).

10. Which of the following statements is incorrect?

• (a) One gram atom of carbon contains Avogadro’s number of atoms.
• (b) One mole of oxygen gas contains Avogadro’s number of molecules.
• (c) One mole of hydrogen contains Avogadro’s number of atoms.
• (d) One mole of electrons stands for \( 6.02 \times 10^{23} \) electrons.

Correct Answer: (c)

Explanation: Statement (c) is incorrect because one mole of hydrogen gas (\( H_{2} \)) contains \( 2 \times N_{A} \) atoms. One mole of hydrogen *atoms* contains Avogadro’s number of atoms.

11. The total number of electrons present in 18 mL of water (density of water is 1 g \( mL^{-1} \)) is:

• (a) \( 6.02 \times 10^{23} \)
• (b) \( 6.02 \times 10^{22} \)
• (c) \( 6.02 \times 10^{24} \)
• (d) \( 6.02 \times 10^{25} \)

Correct Answer: (c)

Explanation: 18 mL \( H_{2}O = 18 \) g \( H_{2}O = 1 \) mole = \( 6.02 \times 10^{23} \) molecules. Since there are 10 electrons in one molecule of \( H_{2}O \), total electrons = \( 10 \times 6.02 \times 10^{23} = 6.02 \times 10^{24} \).

12. A metal oxide has the formula \( M_{2}O_{3} \). It can be reduced by \( H_{2} \) to give free metal and water. 0.1596 g of \( M_{2}O_{3} \) required 6 mg of \( H_{2} \) for complete reduction. The atomic mass of the metal is:

• (a) 27.9
• (b) 79.8
• (c) 55.8
• (d) 159.8

Correct Answer: (c)

Explanation: The reduction reaction is: \( M_{2}O_{3} + 3H_{2} \rightarrow 2M + 3H_{2}O \). Given 0.006 g \( H_{2} \) reduces 0.1596 g oxide, 6 g \( H_{2} \) (standard molar mass for 3 moles) will reduce 159.6 g of \( M_{2}O_{3} \). If \( x \) is the atomic mass, then \( 2x + 48 = 159.6 \Rightarrow 2x = 111.6 \Rightarrow x = 55.8 \).

13. The density of a liquid is 1.2 g/mL. There are 35 drops in 2 mL. The number of molecules in 1 drop is (molecular weight of liquid = 70):

• (a) \( \frac{1.2}{35} N_{A} \)
• (b) \( (\frac{1}{35})^{2} N_{A} \)
• (c) \( \frac{1.2}{(35)^{2}} N_{A} \)
• (d) \( 1.2 N_{A} \)

Correct Answer: (c)

Explanation: Volume of one drop = \( \frac{2}{35} \) mL. Number of moles in one drop = \( \frac{\text{mass}}{\text{mol. wt.}} = \frac{(2/35) \times 1.2}{70} = \frac{1.2}{(35)^{2}} \). Total molecules = \( \frac{1.2}{(35)^{2}} \times N_{A} \).

14. How many millilitres (mL) of 1 M \( H_{2}SO_{4} \) solution is required to neutralise 10 mL of 1 M NaOH solution?

• (a) 2.5 mL
• (b) 5.0 mL
• (c) 10.0 mL
• (d) 20.0 mL

Correct Answer: (b)

Explanation: 1 M \( H_{2}SO_{4} = 2 N \) and 1 M NaOH = 1 N. Using \( N_{1}V_{1} = N_{2}V_{2} \): \( 2 \times V_{1} = 1 \times 10 \Rightarrow V_{1} = 5 \) mL.

15. In a solution, the concentration of \( CaCl_{2} \) is 5 M and concentration of \( MgCl_{2} \) is 5 m. If the specific gravity of the solution is 1.05, the concentration of \( Cl^{-} \) in the solution is:

• (a) 10 M
• (b) 20 M
• (c) 18.5 M
• (d) 17.12 M

Correct Answer: (d)

Explanation: Converting molality to molarity for \( MgCl_{2} \): \( M = 3.56 \). Total \( [Cl^{-}] = (5 \times 2) + (3.56 \times 2) = 17.12 \) M.

16. To 100 mL of 5 M NaOH solution (density 1.2 g/mL) were added 200 mL of another NaOH solution which has a density of 1.5 g/mL and contains 20 mass percent of NaOH. What will be the volume of the gas (at STP) in litres liberated when aluminium reacts with this (final) solution?

• (a) 67.2 L
• (b) 89.6 L
• (c) 44.8 L
• (d) 22.4 L

Correct Answer: (a)

Explanation: Total moles of NaOH = 2.0. From the reaction \( Al + NaOH + H_{2}O \rightarrow NaAlO_{2} + 3/2 H_{2} \), 2 moles of NaOH produce 3 moles of \( H_{2} \). Volume of \( H_{2} \) at STP = \( 3 \times 22.4 = 67.2 \) L.

17. Two oxides of a metal contain 27.6% and 30.0% of oxygen respectively. If the formula of the first oxide is \( M_{3}O_{4} \) then second one is:

• (a) \( MO_{2} \)
• (b) \( M_{2}O \)
• (c) \( M_{2}O_{3} \)
• (d) \( M_{3}O_{2} \)

Correct Answer: (c)

Explanation: Calculations show the ratio of M:O in the second oxide is 2:3, making the formula \( M_{2}O_{3} \).

18. The oxygen-carrying protein known as haemoglobin is 0.335% Fe by mass and contains four Fe atoms per haemoglobin molecule. Calculate the molecular weight of this protein.

• (a) \( 66 g~mol^{-1} \)
• (b) \( 66.6 g~mol^{-1} \)
• (c) \( 6.6683 \times 10^{4} g~mol^{-1} \)
• (d) \( 666 g~mol^{-1} \)

Correct Answer: (c)

Explanation: \( \%Fe = \frac{4 \times \text{At. wt. of Fe}}{\text{Mol. wt. of haemoglobin}} \times 100 = 0.335 \). Molecular weight = \( \frac{400 \times 55.847}{0.335} = 66683 \).

19. A 400 mg iron capsule contains 100 mg of ferrous fumarate, \( (CHCOO)_{2}Fe \). The percentage of iron present in it, is approximately:

• (a) 33%
• (b) 25%
• (c) 14%
• (d) 8%

Correct Answer: (d)

Explanation: Fe in 100 mg of ferrous fumarate is 32.9 mg. Percentage of Fe in the 400 mg capsule = \( \frac{32.9}{400} \times 100 = 8.2\% \).

20. Bromine is prepared commercially by the reaction: \( 2Br^{-}_{(aq)} + Cl_{2(aq)} \rightarrow 2Cl^{-}_{(aq)} + Br_{2(aq)} \). Suppose we have 50.0 mL of 0.060 M solution of NaBr. What volume of 0.05 M solution of \( Cl_{2} \) is needed to react completely with the \( Br^{-} \)?

• (a) 30 mL
• (b) 40 mL
• (c) 20 mL
• (d) 60 mL

Correct Answer: (a)

Explanation: 0.003 mol of \( Br^{-} \) reacts with 0.0015 mol of \( Cl_{2} \). Volume of 0.05 M \( Cl_{2} \) solution = \( \frac{1000}{0.05} \times 0.0015 = 30 \) mL.

21. The largest number of atoms are present in:

• (a) 5 g of \( NH_{3} \)
• (b) 11 g of \( CO_{2} \)
• (c) 8 g of \( SO_{2} \)
• (d) 4 g of \( H_{2} \)

Correct Answer: (d)

Explanation: Number of atoms in (a) \( 5 g NH_{3} = 1.176 N_{A} \); (b) \( 11 g CO_{2} = 0.75 N_{A} \); (c) \( 8 g SO_{2} = 0.375 N_{A} \); (d) \( 4 g H_{2} = 4.00 N_{A} \). Option (d) has the most.

22. One litre of oxygen at STP is made to react with three litres of carbon monoxide at STP. Which one is the limiting reactant?

• (a) CO
• (b) \( O_{2} \)
• (c) \( CO_{2} \)
• (d) None of these.

Correct Answer: (b)

Explanation: In the reaction \( 2CO + O_{2} \rightarrow 2CO_{2} \), 1 vol. \( O_{2} \) reacts with 2 vol. \( CO \). Therefore, \( O_{2} \) is the limiting reagent when 1 L \( O_{2} \) and 3 L CO are taken.

23. In Haber process 30 L of dihydrogen and 30 L of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of the gaseous mixture in the end?

• (a) 20 L \( NH_{3} \), 25 L \( N_{2} \) and 20 L \( H_{2} \)
• (b) 10 L \( NH_{3} \), 25 L \( N_{2} \) and 15 L \( H_{2} \)
• (c) 20 L \( NH_{3} \), 10 L \( N_{2} \) and 30 L \( H_{2} \)
• (d) 20 L \( NH_{3} \), 25 L \( N_{2} \) and 15 L \( H_{2} \)

Correct Answer: (b)

Explanation: 10 L of \( NH_{3} \) is formed (50% yield). This consumes 5 L \( N_{2} \) and 15 L \( H_{2} \). The end mixture contains 25 L \( N_{2} \), 15 L \( H_{2} \), and 10 L \( NH_{3} \).

24. Hydrogen reacts with nitrogen to form ammonia as, \( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \). The amount of ammonia produced if 200.0 g of \( H_{2} \) reacts is:

• (a) 1133.3 g
• (b) 11333 g
• (c) 1032.2 g
• (d) 8692.6 g

Correct Answer: (a)

Explanation: From stoichiometry, 6 g \( H_{2} \equiv 34 \) g of \( NH_{3} \). Thus, 200 g \( H_{2} = \frac{34}{6} \times 200 = 1133.3 \) g.

25. 20 cc of \( CO_{2} \) are passed over red hot coke. The volume of CO evolved is:

• (a) 10 cc
• (b) 20 cc
• (c) 30 cc
• (d) 40 cc

Correct Answer: (d)

Explanation: In the reaction \( CO_{2} + C \rightarrow 2CO \), 1 cc of \( CO_{2} \) gives 2 cc of CO. Thus, 20 cc \( CO_{2} \) gives 40 cc of CO.

26. 0.30 g of an organic compound containing C, H and O on combustion yielded 0.44 g \( CO_{2} \) and 0.18 g \( H_{2}O \). If 1 mole of compound weighs 60 g, then molecular formula of the compound is:

• (a) \( C_{2}H_{4}O_{2} \)
• (b) \( CH_{2}O \)
• (c) \( C_{3}H_{8}O \)
• (d) \( C_{4}H_{12} \)

Correct Answer: (a)

Explanation: Calculation gives an empirical formula of \( CH_{2}O \) (mass 30). Since the molecular mass is 60, \( n = 2 \), making the molecular formula \( C_{2}H_{4}O_{2} \).

27. How much water is needed to dilute 10 mL of 10 N hydrochloric acid to make it exactly decinormal (0.1 N)?

• (a) 990 mL
• (b) 1000 mL
• (c) 1010 mL
• (d) 100 mL

Correct Answer: (a)

Explanation: Using \( N_{1}V_{1} = N_{2}V_{2} \): \( 10 \times 10 = 0.1 \times V_{2} \Rightarrow V_{2} = 1000 \) mL. Volume of water to add = \( 1000 – 10 = 990 \) mL.

28. Two oxides of a certain metal were separately heated in a current of hydrogen until constant weights were obtained. 2 g of each oxide gave, respectively 0.2517 g and 0.4526 g of water. This observation illustrates:

• (a) law of conservation of mass
• (b) law of constant proportions
• (c) law of multiple proportions
• (d) law of reciprocal proportions.

Correct Answer: (c)

Explanation: The weights of oxygen combining with a constant weight of metal are in a simple ratio of 1:2, illustrating the law of multiple proportions.

29. Which one of the following represents smallest quantity?

• (a) 1850 ng
• (b) \( 1.85 \times 10^{-4} \) g
• (c) \( 1.85 \times 10^{3} \mu g \)
• (d) \( 1.85 \times 10^{-6} kg \)

Correct Answer: (a)

Explanation: Conversion to grams: (a) \( 1.85 \times 10^{-6} \) g; (b) \( 1.85 \times 10^{-4} \) g; (c) \( 1.85 \times 10^{-3} \) g; (d) \( 1.85 \times 10^{-3} \) g. Option (a) is the smallest.

30. 10 g piece of a marble was put into excess of dil. HCl. When the reaction was complete, 1120 \( cm^{3} \) of \( CO_{2} \) was obtained at STP. The percentage of \( CaCO_{3} \) in the marble is:

• (a) 25%
• (b) 50%
• (c) 75%
• (d) 10%

Correct Answer: (b)

Explanation: 1120 cc of \( CO_{2} \) is obtained from 5 g of pure \( CaCO_{3} \). Percentage of \( CaCO_{3} = \frac{5}{10} \times 100 = 50\% \).

31. 50.0 kg of \( N_{2(g)} \) and 10.0 kg of \( H_{2(g)} \) are mixed to produce \( NH_{3(g)} \). Identify the limiting reagent.

• (a) Nitrogen
• (b) Hydrogen
• (c) Can not be predicted
• (d) None of these.

Correct Answer: (b)

Explanation: 1785.7 mol of \( N_{2} \) require 5357.1 mol of \( H_{2} \). Since only 4960.3 mol of \( H_{2} \) are available, \( H_{2} \) is the limiting reagent.

32. How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl?

• (a) 0.339
• (b) 0.011
• (c) 0.029
• (d) 0.044

Correct Answer: (c)

Explanation: PbO is the limiting reagent (0.029 mol). Therefore, 0.029 mol of \( PbCl_{2} \) is produced.

33. 10 g of hydrofluoric acid gas occupies 5.6 L of volume at STP. If the empirical formula is HF, then its molecular formula in the gaseous state will be:

• (a) HF
• (b) \( H_{2}F_{2} \)
• (c) \( H_{3}F_{3} \)
• (d) \( H_{4}F_{4} \)

Correct Answer: (b)

Explanation: Molecular mass = \( \frac{10}{5.6} \times 22.4 = 40 \). E.F. mass of HF = 20. \( n = 2 \), thus the molecular formula is \( H_{2}F_{2} \).

34. If density of 3 M of NaCl is 1.25 g \( mL^{-1} \), the molality of the solution will be:

• (a) 3 m
• (b) 2.50 m
• (c) 1.75 m
• (d) 2.79 m

Correct Answer: (d)

Explanation: Molality = \( \frac{3}{1.0745} = 2.79 \) m.

35. The following data are obtained when carbon and dioxygen react to form different compounds: (i) 12g C + 16g O (ii) 12g C + 32g O. Which law is obeyed?

• (a) Law of definite proportions
• (b) Law of multiple proportions
• (c) Gay Lussac’s law
• (d) Avogadro law.

Correct Answer: (b)

Explanation: The masses of oxygen (16 g and 32 g) that combine with a fixed mass of carbon (12 g) are in a simple ratio (1:2), obeying the law of multiple proportions.

36. How many molecules of \( CO_{2} \) are formed when one milligram of 100% pure \( CaCO_{3} \) is treated with excess hydrochloric acid?

• (a) \( 6.023 \times 10^{23} \)
• (b) \( 6.023 \times 10^{21} \)
• (c) \( 6.023 \times 10^{20} \)
• (d) \( 6.023 \times 10^{18} \)

Correct Answer: (d)

Explanation: 100 g \( CaCO_{3} \) gives \( 6.023 \times 10^{23} \) molecules. Thus, \( 10^{-3} \) g gives \( \frac{6.023 \times 10^{23}}{100} \times 10^{-3} = 6.023 \times 10^{18} \).

37. An unknown element forms an oxide. What will be the equivalent weight of the element if the oxygen content is 20% by weight?

• (a) 16
• (b) 32
• (c) 8
• (d) 64

Correct Answer: (b)

Explanation: 20 g of oxygen combines with 80 g of metal. Therefore, 8 g of oxygen combines with \( \frac{80}{20} \times 8 = 32 \) g of metal. Equivalent weight = 32.

38. The amount of water (g) produced by the combustion of 16 g of methane is:

• (a) 16 g
• (b) 36 g
• (c) 18 g
• (d) 32 g

Correct Answer: (b)

Explanation: 1 mole of \( CH_{4} \) (16 g) gives 2 moles of \( H_{2}O \) (36 g).

39. How many moles of methane are required to produce 22 g of \( CO_{2(g)} \) after combustion?

• (a) 0.5 mol
• (b) 1 mol
• (c) 0.25 mol
• (d) 1.5 mol

Correct Answer: (a)

Explanation: 1 mol \( CO_{2} \) (44 g) is obtained from 1 mol \( CH_{4} \). Thus, 22 g \( CO_{2} \) is obtained from 0.5 mol \( CH_{4} \).

40. Calculate the amount of \( CO_{2(g)} \) produced by combustion of 32 g of oxygen.

• (a) 22 g
• (b) 240 g
• (c) 21 g
• (d) 32 g

Correct Answer: (a)

Explanation: 2 moles of \( O_{2} \) (64 g) give 1 mole of \( CO_{2} \) (44 g). Thus, 1 mole of \( O_{2} \) (32 g) gives 0.5 mole of \( CO_{2} \) (22 g).

41. 2 g of metal carbonate is neutralized completely by 100 mL of 0.1 N HCl. The equivalent weight of metal carbonate is:

• (a) 50
• (b) 100
• (c) 150
• (d) 200

Correct Answer: (d)

Explanation: Gram equivalents of HCl = 0.01. Thus, gram equivalents of carbonate \( = \frac{2}{E} = 0.01 \Rightarrow E = 200 \).

42. The volume of carbon dioxide gas evolved at S.T.P. by heating 7.3 g of \( Mg(HCO_{3})_{2} \) will be:

• (a) 2240 mL
• (b) 1120 mL
• (c) 2340 mL
• (d) 2000 mL

Correct Answer: (a)

Explanation: 146 g of \( Mg(HCO_{3})_{2} \) gives 44.8 L of \( CO_{2} \). Thus, 7.3 g gives 2.24 L or 2240 mL of \( CO_{2} \).

43. In a compound C, H and N are present in 9 : 1 : 3.5 by weight. If molecular weight is 108, the molecular formula is:

• (a) \( C_{2}H_{6}N_{2} \)
• (b) \( C_{3}H_{4}N \)
• (c) \( C_{6}H_{8}N_{2} \)
• (d) \( C_{9}H_{12}N_{3} \)

Correct Answer: (c)

Explanation: The empirical formula is \( C_{3}H_{4}N \) (mass 54). Since the molecular mass is 108, \( n = 2 \), making the formula \( C_{6}H_{8}N_{2} \).

44. A compound with empirical formula \( CH_{2}O \) has a vapour density of 30. Its molecular formula is:

• (a) \( C_{2}H_{2}O_{2} \)
• (b) \( C_{2}H_{4}O_{2} \)
• (c) \( C_{3}H_{6}O_{3} \)
• (d) \( C_{6}H_{12}O_{6} \)

Correct Answer: (b)

Explanation: Molecular weight = \( 2 \times 30 = 60 \). \( n = \frac{60}{30} = 2 \). Molecular formula = \( (CH_{2}O)_{2} = C_{2}H_{4}O_{2} \).

45. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/mL. The molarity of the solution is:

• (a) 1.78 M
• (b) 2.00 M
• (c) 2.05 M
• (d) 2.22 M

Correct Answer: (c)

Explanation: Total solution mass = 1120 g. Volume = \( \frac{1120}{1.15} = 973.91 \) mL. Moles of urea = 2. Molarity = \( \frac{2 \times 1000}{973.91} = 2.05 \) M.