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 weight of the substance will be:

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Q. 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 $80^{\circ}C$ is $750\, mm$. The molecular weight of the substance will be:

Solutions

Solution:

$\frac{P_{0}-P_{s}}{P_{s}}=\frac{w\times M}{m\times W}$
$\frac{10}{\left(750-10\right)}=\frac{2\times78}{m\times78}$
$\therefore m=148$ ;
NOTE: [m comes 150 if formula $\frac{P_{0}-P_{s}}{P_{0}}=\frac{w \times M}{m\times W}$ is used. But this is only for dilute solutions]