The vapour pressure of 30 % (w/v) aqueous sol | vedclass

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

Question

(A) The relative lowering of vapour pressure is given by $\frac{P^{\circ} - P_{s}}{P_{s}} = \frac{n_{solute}}{n_{solvent}}$.Given: $30 \% \ (w/v)$ sol

Vedclass · Accepted Answer

$23$

Vedclass · Answer

(A) The relative lowering of vapour pressure is given by $\frac{P^{\circ} - P_{s}}{P_{s}} = \frac{n_{solute}}{n_{solvent}}$.Given: $30 \% \ (w/v)$ solution means $30 \ g$ of glucose in $100 \ mL$ of solution.Mass of solution $= 	ext{density} 	imes 	ext{volume} = 1.2 \ g \ cm^{-3} 	imes 100 \ mL = 120 \ g$.Mass of solvent (water) $= 120 \ g - 30 \ g = 90 \ g$.Moles of glucose $(n_{solute})$ $= \frac{30 \ g}{180 \ g \ mol^{-1}} = \frac{1}{6} \ mol \approx 0.1667 \ mol$.Moles of water $(n_{solvent})$ $= \frac{90 \ g}{18 \ g \ mol^{-1}} = 5 \ mol$.Using the formula $\frac{24 - P_{s}}{P_{s}} = \frac{0.1667}{5} = 0.03334$.$24 - P_{s} = 0.03334 \ P_{s} \implies 24 = 1.03334 \ P_{s}$.$P_{s} = \frac{24}{1.03334} \approx 23.22 \ mm \ Hg$. The closest integer value is $23 \ mm \ Hg$.

Similar Questions

Assertion $:-$ Vapour pressure of a liquid increases on increasing surface area.
Reason $:-$ Rate of vaporisation of a liquid is directly proportional to surface area.

On adding a solute to a solvent having a vapour pressure of $0.80 \ atm$,the vapour pressure reduces to $0.60 \ atm$. The mole fraction of the solute is:

The relative lowering of vapor pressure is equal to the mole fraction of the non-volatile solute. This statement was given by whom?

What is the molar mass of a solute when $2.3 \ g$ of a non-volatile solute is dissolved in $46 \ g$ of benzene at $30^{\circ} C$? (Relative lowering of vapour pressure is $0.06$ and the molar mass of benzene is $78 \ g \ mol^{-1}$)

The vapour pressure will be lowest for

Vedclass Test Series

Exam Paper Generator

Online Exam Module