The vapour pressure of pure CCl_4 (molar mass | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given, vapour pressure of pure $CCl_4$ at $25^{\circ} C$ $(p^{\circ}_{CCl_4}) = 115.0 \ torr$. Vapour pressure of pure $SnCl_4$ at $25^{\circ} C$

Vedclass · Accepted Answer

$185.85$

Vedclass · Answer

(A) Given, vapour pressure of pure $CCl_4$ at $25^{\circ} C$ $(p^{\circ}_{CCl_4}) = 115.0 \ torr$. Vapour pressure of pure $SnCl_4$ at $25^{\circ} C$ $(p^{\circ}_{SnCl_4}) = 238.0 \ torr$. Moles of $CCl_4$ $(n_{CCl_4}) = \frac{10 \ g}{154 \ g \ mol^{-1}} \approx 0.0649 \ mol$. Moles of $SnCl_4$ $(n_{SnCl_4}) = \frac{15 \ g}{170 \ g \ mol^{-1}} \approx 0.0882 \ mol$. Total moles $(n_{total}) = 0.0649 + 0.0882 = 0.1531 \ mol$. Mole fraction of $CCl_4$ $(\chi_{CCl_4}) = \frac{0.0649}{0.1531} \approx 0.424$. Mole fraction of $SnCl_4$ $(\chi_{SnCl_4}) = 1 - 0.424 = 0.576$. Total vapour pressure $(p_{total}) = \chi_{CCl_4} \cdot p^{\circ}_{CCl_4} + \chi_{SnCl_4} \cdot p^{\circ}_{SnCl_4}$. $p_{total} = (0.424 	imes 115.0) + (0.576 	imes 238.0) = 48.76 + 137.09 = 185.85 \ torr$. Thus, the correct option is $A$.

Similar Questions

Lowering of vapour pressure due to a solute in $1 \, molal$ aqueous solution at $100 \, ^\circ C$ is ........ $torr$.

Derive an equation for a solution that shows the relation between total pressure and the mole fraction of a volatile solute and a volatile solvent,and explain it by plotting a graph.

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in $1 \ kg$ of water is $0.002$. The molality of the solution is (in $m$)

$5 \ cm^3$ of acetone is added to $100 \ cm^3$ of water. The vapour pressure of water over the solution:

Vedclass Test Series

Exam Paper Generator

Online Exam Module