chemistry

Let the total number of moles of the solution be 1.

Hence, the number of moles of benzene are $x$.

The molar mass of benzene is $78$ g/mol.

The mass of benzene is $xmol×78g/mol=78xg$. The density of benzene is $0.87g/ml$.

The mass of benzene is $xmol×78g/mol=78xg$. The density of benzene is $0.87g/ml$.

Hence, the volume of liquid benzene is $0.877g/ml78xg =88.9ml=0.0889L$.

In the vapor phase, the volume of benzene will be $2750×0.0889L=244.5L$. The vapor pressure of benzene is

$P=VnRT =244.5xLxmol×0.0821L⋅atm/mol/K×293K =0.0984atm$

Convert the unit of pressure from atm to torr using the relation $1atm=760torr$.

$0.0984atm×1atm760torr =74.8torr$.

The mole fraction of benzene in the vapor phase is $74.8torr46.0torr =0.62$.

Answer verified by Toppr

(a) external pressure is reduced to 1 atm in single step

(b) external pressure is reduced to 5 atm in 1st operation and then 1 atm in next step

(c) gas is allowed to expand into an evacuated space of 10 litre

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[Equilibrium vapour pressure for $H_{2}O$ at $21_{∘}C≃19torr$].[R=0.082 liter atm mole $_{−1}$deg$_{−1},$molecular weight $H_{2}O=18$]

[Write your answer to the nearest integer].

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$H_{2}+Cl_{2}→2HCl$

How many grams of chlorine gas are required to produce $3L$ of $HCl$ gas at a pressure of $2atm$ and a temperature of $19_{0}C$?

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