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When 0.1 m CH₃COOH is present in a solvent it shows elevation in boiling point of 0.75 $° C$ . Acid dissociation constant will be -(K_b=5 K Kg mol^-1) (1M =1m)

5 X 10^-2

5 X 10^-3

10^-2

10^-3

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Explanation

${ΔT}_{b} = i K_{b} m; 0.75 = i \times 5 \times 0.1 i = 1.5; α = \frac{1.5 - 1}{2 - 1} = 0.5 K_{a} = \frac{α^{2} C}{1 - α} = \frac{{(0.5)}^{2} \times 0.1}{1 - 0.5} = 5 \times 10^{- 2}$

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Which of the following shows positive deviation from Raoult's law:

(i) Chloroform and acetone

(ii) Carbon disulphide and acetone

(iii) Ethanol and Acetone

(iv) Phenol and Aniline

i and ii

ii and iii

i and iV

ii and iv

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A non-volatile solute is dissolved in a water. If its degree of association is 50%. What will be its freezing point when boiling point is 100.52 $° C$ . [K_b=0.52 K kg mol^-1, K_f=1.86K kg mol^-1]

-0.52 K

-0.52 $° C$

-1.86 $° C$

275.01 K

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Explanation

For a non-volatile solute, the freezing point depression (Î”Tf) is given by Î”Tf = Kf * m, where Kf is the cryoscopic constant and m is the molality of the solution. Given that the degree of association is 50%, the van't Hoff factor (i) is 2. Substituting the values, Î”Tf = 1.86 * (100.52/100) * 2 = 3.72Â°C. Therefore, the freezing point will be 100.52 - 3.72 = 96.8Â°C or -1.86Â°C.

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When 36 g of a solute having the emperical formula CH₂O is dissolved in 1.2 kg of water, the solution freezes at -0.93 $° C$ . What is the molecular formula of solute (K_f=1.86 kg K mol^-1)

C₂H₄O

C₂H₂O₂

C₂H₄O₃

C₂H₄O₂

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Explanation

$∆ T_{f} = {ik}_{f} m 0.93 = 1 \times 1.86 \times \frac{36}{1.2 M}; M = 60$

m is 60 gram , so 4th is right ans

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How much oxygen is dissolved in 100ml water at 298K if partial pressure of oxygen is 0.5 atm and K_H=1.4 X 10^-3 M/atm ?

22.4 mg

22.4 g

2.24 g

2.24 mg

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Explanation

According to henry's law

S=K_H X p (S=conc. of O₂ dissolved)

$S = 1.4 \times 10^{- 3} \times 0.5 = 7 \times 10^{- 4} mol / L = \frac{w / M_{w}}{V_{L}} \Rightarrow w = 7 \times 10^{- 4} \times M_{w} \times V_{L} = 7 \times 10^{- 4} \times 32 \times 0.1 = 22.4 \times 10^{- 4} g = 2.24 mg$

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A solute x when dissolved in solvent associates to form a pentamer. The value of van't hoff factor (i) for the solute will be:-

0.5

0.2

0.1

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Explanation

$5 x \to x_{5} DOA, \infty \frac{1 - i}{1 - \frac{1}{n}} \Rightarrow 1 = \frac{1 - i}{1 - \frac{1}{5}} \Rightarrow \frac{4}{5} = 1 - i \Rightarrow i = 1 - \frac{4}{5} = \frac{1}{5} = 0.2$

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What amount of CaCl₂(i=2.47) is dissolved in 2L water so that its osmotic pressure is 0.5 atm at 27 $° C$ :-

13.42 g

19.24 g

8.834 g

1.820 g

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Explanation

$π = iCRT = i \times \frac{n}{V} \times RT n = \frac{π \times V}{i \times R \times T} = \frac{0.5 \times 2}{2.47 \times 0.0821 \times 300} = 0.0164 mol ∴ w = 0.0164 \times 111 g = 1.820 g$

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One litre aqueous solution of sucrose (molar mass = 342 g/mol) weighing 1015 g is found to record an osmotic pressure of 4.8 atm at 293K. What is the molarity of the sucrose solution ? (R= 0.0821 L atm K^-1 mol^-1) :

0.84

0.21

0.42

0.63

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The vapour pressure of water at room temperature is 30 mm of Hg. If the mole fraction of the water is 0.9, the vapour pressure of solution will be

30 mm of Hg

24 mm of HG

21 mm of Hg

27 mm of Hg

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Explanation

$\frac{P - P_{s}}{P} = \frac{n}{n + N} = 0.1 \frac{30 - P_{s}}{30} = 0.1 P_{s} = 27 mm of Hg$

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The Van't Haff factor of a 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCl is:

0.95

0.97

0.94

0.96

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Explanation

For KCl, i=1+x

1.95=1+ x

x=0.95

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