In 108 , g of water,18 , g of a non-volatile | vedclass

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

Question

(D) The relative lowering of vapour pressure is given by the formula: $\frac{p_0 - p_s}{p_0} = \chi_2 = \frac{n_2}{n_1 + n_2}$.Given: $p_0 = 760 \, mm

Vedclass · Accepted Answer

$228$

Vedclass · Answer

(D) The relative lowering of vapour pressure is given by the formula: $\frac{p_0 - p_s}{p_0} = \chi_2 = \frac{n_2}{n_1 + n_2}$.Given: $p_0 = 760 \, mm \, Hg$ (at $100^{\circ} C$),$p_s = 750 \, mm \, Hg$,$w_2 = 18 \, g$,$w_1 = 108 \, g$,$M_1 = 18 \, g \, mol^{-1}$.$n_1 = \frac{108}{18} = 6 \, mol$.$n_2 = \frac{18}{M_2}$.Using the formula $\frac{760 - 750}{760} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (since $n_2$ is small compared to $n_1$):$\frac{10}{760} = \frac{18/M_2}{6}$.$\frac{1}{76} = \frac{3}{M_2}$.$M_2 = 3 	imes 76 = 228 \, g \, mol^{-1}$.

In $108 \, g$ of water,$18 \, g$ of a non-volatile compound is dissolved. At $100^{\circ} C$,the vapour pressure of the solution is $750 \, mm \, Hg$. Assuming that the compound does not undergo association or dissociation,the molar mass of the compound in $g \, mol^{-1}$ is

Similar Questions

The vapour pressure of pure liquids $A$ and $B$ are $450$ $mm$ $Hg$ and $700$ $mm$ $Hg$ respectively,at $350$ $K$. Find out the composition of the liquid mixture if the total vapour pressure is $600$ $mm$ $Hg$. Also,find the composition of the vapour phase.

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 $m$)

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

Calculate the relative lowering of vapour pressure if the vapour pressure of benzene and vapour pressure of solution of non-volatile solute in benzene are $640 \ mmHg$ and $590 \ mmHg$ respectively at the same temperature.

What is the vapour pressure of a solution containing $0.1 \text{ mol}$ of non-volatile solute dissolved in $16.2 \text{ g}$ water (in $\text{ mmHg}$)? ($P_1^0 = 24 \text{ mmHg}$, molar mass of water $18 \text{ g mol}^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module