If 8 ; g of a non-electrolyte solute is disso | vedclass

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

Question

(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \fr

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(B) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by:$\frac{P_{A}^{0} - P_{s}}{P_{A}^{0}} = \frac{n_{B}}{n_{A} + n_{B}} \approx \frac{n_{B}}{n_{A}}$ (for dilute solutions).Given:$w_{B} = 8 \; g$,$W_{A} = 114 \; g$,$M_{A} = 114 \; g \; mol^{-1}$.Since the vapour pressure is reduced to $80 \%$,$P_{s} = 0.8 P_{A}^{0}$.Therefore,$\frac{P_{A}^{0} - 0.8 P_{A}^{0}}{P_{A}^{0}} = \frac{0.2 P_{A}^{0}}{P_{A}^{0}} = 0.2$.Using the formula $\frac{\Delta P}{P_{A}^{0}} = \frac{w_{B} 	imes M_{A}}{m_{B} 	imes W_{A}}$:$0.2 = \frac{8 	imes 114}{m_{B} 	imes 114}$.$0.2 = \frac{8}{m_{B}}$.$m_{B} = \frac{8}{0.2} = 40 \; g \; mol^{-1}$.

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.
Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.

Similar Questions

If the vapour pressure of a solution containing a non-volatile solute is $2\%$ less than the vapour pressure of pure water,find the molality of the solution.

If two substances $A$ and $B$ have $P_{A}^{\circ} : P_{B}^{\circ} = 1 : 2$ and have mole fraction in solution in the ratio $1 : 2$,then mole fraction of $A$ in vapour phase is:-

Vapour pressure of $CCl_{4}$ at $25\,^{\circ} C$ is $143\, mm\, Hg$. $0.5\, g$ of a non-volatile solute (mol. wt. $65$) is dissolved in $100\, mL$ of $CCl_{4}$. Find the vapour pressure of the solution. (Density of $CCl_{4} = 1.58\, g / cm^{3}$)

When $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water,the vapour pressure of the aqueous solution at $100 \ ^\circ C$ is .......... $torr$.

At $25\,^oC$,the vapour pressure of $CCl_4$ is $143\,mm\,Hg$. If $0.5\,g$ of a non-volatile solute (molar mass $= 65\,g/mol$) is dissolved in $100\,mL$ of $CCl_4$,what will be the vapour pressure of the solution? (Density of $CCl_4 = 1.58\,g/cm^3$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If $8 \; g$ of a non-electrolyte solute is dissolved in $114 \; g$ of $n$-octane to reduce its vapour pressure to $80 \%$,the molar mass (in $g \; mol^{-1}$) of the solute is.Given that the molar mass of $n$-octane is $114 \; g \; mol^{-1}$.