When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 99.652 KPa. The distillate contained 1.27 gm of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is 85.140 KPa at 95.2°C ?

Question

When a liquid that is immiscible with water was steam distilled at 95.2°C at a total pressure of 99.652 KPa. The distillate contained 1.27 gm of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is 85.140 KPa at 95.2°C ? (A) 134.1 gm mol-1 (B) 105.74 gm mol-1 (C) 99.65 gm mol-1 (D) 18 gm mol-1

Tardigrade Admin · Accepted Answer

Total pressure, p text total =99.652 kPa pB =p text water =85.140 kPa p text liquid =pA=(99.652-85.140) kPa =14.512 kPa and (mA/mB)= (1.27/1) or (mA/mB)=(pA MA/pB MB) or, MA =((mA/mB))((pB MB/pA)) =1.27 ×((85.140 kPa × 18 g mol -1/14.512 kPa )) ≃ 134.1 g mol -1

$134.1 g m m o l^{- 1}$

$105.74 g m m o l^{- 1}$

$99.65 g m m o l^{- 1}$

$18 g m m o l^{- 1}$

Q. When a liquid that is immiscible with water was steam distilled at 95.2∘C at a total pressure of 99.652KPa. The distillate contained 1.27gm of the liquid per gram of water. What will be the molar mass of the liquid if the vapour pressure of water is 85.140KPa at 95.2∘C ?

134.1gmmol−1

105.74gmmol−1

99.65gmmol−1

18gmmol−1

Total pressure, ptotal ​=99.652kPa pB​=pwater ​=85.140kPa pliquid ​=pA​=(99.652−85.140)kPa =14.512kPa and mB​mA​​=11.27​ or mB​mA​​=pB​MB​pA​MA​​ or, MA​=(mB​mA​​)(pA​pB​MB​​) =1.27×(14.512kPa85.140kPa×18gmol−1​) ≃134.1gmol−1

$134.1 g m m o l^{- 1}$

$105.74 g m m o l^{- 1}$

$99.65 g m m o l^{- 1}$

$18 g m m o l^{- 1}$