1. Tardigrade
  2. Question
  3. Chemistry
  4. The vapour pressure of benzene at 80° C is lowered by 10 mm by dissolving 2 g of a non-volatile substance in 78 g of benzene. The vapour pressure of pure benzene at 80° C is 750 mm. The molecular mass of the substance will be :

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Q. The vapour pressure of benzene at $80^{\circ} C$ is lowered by $10 \,mm$ by dissolving $2\, g$ of a non-volatile substance in $78 \,g$ of benzene. The vapour pressure of pure benzene at $80^{\circ} C$ is $750\, mm$. The molecular mass of the substance will be :

Solution:

$\frac{ P ^{\circ}- P _{ s }}{ P ^{\circ}}=X_{\text {Benzene }}$
$\Rightarrow \frac{750-740}{750}=\frac{2 / M }{\frac{2}{ M }+\frac{78}{78}}$
On Solving, $M=148$