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The vapour pressure of acetone at $25^{\circ}$ is 192 torr. When $1.5 \mathrm{~g}$ of non-volatile substance was dissolved in $100 \mathrm{~g}$ of acetone at $25^{\circ} \mathrm{C}$ its vapour pressure was 189 torr. The molar mass $\left(\mathrm{gmol}^{-1}\right)$ of substance is
Option: 1
$\mathrm{ 42=m }$

Option: 2
$\mathrm{ 92=m }$

Option: 3
$\mathrm{ 55.6=m }$

Option: 4
$\mathrm{102=m }$

Answers (1)

best_answer

$\mathrm{\frac{192-189}{192}=\frac{1.5 / m}{100 / 58} }$
$\mathrm{m=55.6 }$ .

Posted by

Kuldeep Maurya

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