The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass mass 78 g mol -1 ). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Question

The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass mass 78 g mol -1 ). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance? (A) 150 g mol-1 (B) 200 g mol-1 (C) 100 g mol-1 (D) 170 g mol -1

Tardigrade Admin · Accepted Answer

(p°-P1/P°) = (w2 × M1/M2× w1) (0.850 bar -0.845 bar/0.850 bar) = (0.5 g × 78 g mol-1/M2 × 39 g) ∴ M2 = 170 g mol-1

$150 g m o l^{- 1}$

$200 g m o l^{- 1}$

$100 g m o l^{- 1}$

$170 g m o l^{- 1}$

$\frac{p ^{\circ} - P _{1}}{P ^{\circ}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$
$\frac{0.850 ba r - 0.845 ba r}{0.850 ba r}$
$= \frac{0.5 g \times 78 g m o l ^{- 1}}{M _{2} \times 39 g}$
$∴ M_{2} = 170 g m o l^{- 1}$

Q. The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0g of benzene (molar mass mass 78gmol−1 ). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

150gmol−1

200gmol−1

100gmol−1

170gmol−1

P∘p∘−P1​​=M2​×w1​w2​×M1​​ 0.850bar0.850bar−0.845bar​ =M2​×39g0.5g×78gmol−1​ ∴M2​=170gmol−1

$150 g m o l^{- 1}$

$200 g m o l^{- 1}$

$100 g m o l^{- 1}$

$170 g m o l^{- 1}$

$\frac{p ^{\circ} - P _{1}}{P ^{\circ}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$
$\frac{0.850 ba r - 0.845 ba r}{0.850 ba r}$
$= \frac{0.5 g \times 78 g m o l ^{- 1}}{M _{2} \times 39 g}$
$∴ M_{2} = 170 g m o l^{- 1}$