The vapour pressure of pure benzene is 639.70 mm of Hg and the vapour pressure of solution of a solute in benzene at the same temperature is 631.9 mm of Hg. Calculate the molality of the solution.

Question

The vapour pressure of pure benzene is 639.70 mm of Hg and the vapour pressure of solution of a solute in benzene at the same temperature is 631.9 mm of Hg. Calculate the molality of the solution. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

According to Raoult's law: p=p0 chi1 ⇒ 631.9=639.7 chi1 ⇒ chi1=0.9878 ⇒ chi2=0.0122 ⇒ Molality =(0.0122/0.9878 × 78) × 1000=0.158

Q. The vapour pressure of pure benzene is $639.70 mm$ of $H g$ and the vapour pressure of solution of a solute in benzene at the same temperature is $631.9 mm$ of $H g$ . Calculate the molality of the solution.

According to Raoult's law :
$p = p_{0} χ_{1} \Rightarrow 631.9 = 639.7 χ_{1}$
$\Rightarrow χ_{1} = 0.9878 \Rightarrow χ_{2} = 0.0122$
$\Rightarrow$ Molality $= \frac{0.0122}{0.9878 \times 78} \times 1000 = 0.158$

Q. The vapour pressure of pure benzene is 639.70mm of Hg and the vapour pressure of solution of a solute in benzene at the same temperature is 631.9mm of Hg. Calculate the molality of the solution.

According to Raoult's law : p=p0​χ1​⇒631.9=639.7χ1​ ⇒χ1​=0.9878⇒χ2​=0.0122 ⇒ Molality =0.9878×780.0122​×1000=0.158

Q. The vapour pressure of pure benzene is $639.70 mm$ of $H g$ and the vapour pressure of solution of a solute in benzene at the same temperature is $631.9 mm$ of $H g$ . Calculate the molality of the solution.

According to Raoult's law :
$p = p_{0} χ_{1} \Rightarrow 631.9 = 639.7 χ_{1}$
$\Rightarrow χ_{1} = 0.9878 \Rightarrow χ_{2} = 0.0122$
$\Rightarrow$ Molality $= \frac{0.0122}{0.9878 \times 78} \times 1000 = 0.158$