NEET UG CHEMISTRY MCQ: SOLUTIONS

1 Hard

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Analyzing +4 | -1

The freezing point of an aqueous solution of

{KCN}

containing 0.2 mole/kg water was

-0.8^{\circ} \mathrm{C}

. On addition of 0.1 mole of

{Hg}({CN})_2

in the solution containing 1 kg of water, the freezing point of the solution was

-0.6^{\circ} \mathrm{C}

. Assuming that the complex is formed according to the following equation:

{Hg}({CN})_2+{mCN}^{-} \longrightarrow {Hg}({CN})_{{m}+2}^{-{m}}

and

{Hg}({CN})_2

is the limiting reagent, the value of

4 \mathrm{~m}

is:

View answer and explanation

Correct option: A

Option A (8) is correct because the freezing point depression indicates the formation of a complex ion with 2 additional cyanide ions, thus m=2 and 4m=8.

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Consider the solution of ethanol. In it Assertion: Isotonic solutions don't show net phenomenon of osmosis.

Reason: Isotonic solutions have equal osmotic pressure at constant temperature. 0.2 molal aqueous solution of a monobasic acid is 60% ionised. If Kb for water is

0.52\,K\,kg\,mo{l^{ - 1}}

then the boiling point of the solution will be The vapour pressures of pure liquids

A

and

B

are 400 and 600 mmHg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid

B

is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components

A

and

B

in vapour phase, respectively are What is molecular weight of a substance whose 7.0% by mass solution in water freezes at

--{0.93^0}C

{K_f}

for water is

1.86\mathrm{ }K\mathrm{ }kg\mathrm{ }mo{l^{ - 1}}

A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. the vapour pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in mm) at the same temperature will be