JEE Main 2019 (Online) 10th April Morning Slot | Solutions Question 103 | Chemistry

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be. (molar mass of urea = 60 g mol^–1)

0.031 mmHg

0.017 mmHg

0.028 mmHg

0.027 mmHg

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

Molal depression constant for a solvent is 4.0 kg mol^–1. The depression in the freezing point of the solvent for 0.03 mol kg^–1 solution of K₂SO₄ is :
(Assume complete dissociation of the electrolyte)

0.18 K

0.24 K

0.36 K

0.12 K

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl₂ in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L^–1) in solution is :

4 × 10^–4

6 × 10^–2

4 × 10^–2

16 × 10^–4

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is:
(x_M = Mole fraction of 'M' in solution ;
x_N = Mole fraction of 'N' in solution ;
y_M = Mole fraction of 'M' in vapour phase ;
y_N = Mole fraction of 'N' in vapour phase)

$${{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}$$

(x_M – y_M) < (x_N – y_N)

$${{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}$$

$${{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}$$

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