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If 8 g of a non - electrolyte solute is  dissolved in 114 g of n-octane to reduce its vapour  pressure to 80%, the molar mass (in g mol -1 ) of the solute is

[Given that molar mass of n-octane is \mathrm{114\ g \ mol^{-1}}]

Option: 1

60


Option: 2

80


Option: 3

20

 


Option: 4

40


Answers (1)

best_answer

We know that,

Relative lowering of vapour pressure = Mole fraction of solute

\mathrm{\frac{P_{o}-P_{s}}{P_{o}}=\frac{n_{B}}{n_{A}+n_{B}}}

\mathrm{\Rightarrow 0.2=\frac{n_{B}}{n_{A}+n_{B}}}

Assuming that the solution is very dilute

i.e. \mathrm{n_{A}+n_{B} \simeq n_{A}}

We have,

\mathrm{0.2=\frac{n_{B}}{n_{A}}}

\mathrm{\Rightarrow n_{B} = 0.2 \times n_{A}}

\mathrm{\Rightarrow n_{B} = 0.2}

\mathrm{\Rightarrow \frac{8}{M_{B}}=0.2}

\mathrm{\Rightarrow M_{B} =40}

Hence, the correct answer is option 4.

Note : NTA's assumption of very dilute solution is although not correct, but no other option can be arrived at by not neglecting n_{B} with respect to n_{A}.

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