NEET UG CHEMISTRY MCQ: SOLUTIONS

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1 mole of heptane (V.P = 92 mm of Hg) was mixed with 4 moles of octane (V.P = 31 mm of Hg). The V.P of resulting ideal solution is

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Correct option: C

43.2 mm of Hg

Option C (43.2 mm of Hg) is correct because it is calculated using Raoult's law for ideal solutions, considering the mole fractions and vapor pressures of each component.

More Questions from SOLUTIONS

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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