The mole fraction of a solvent in an aqueous | vedclass

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

Question

(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent}

Vedclass · Accepted Answer

$13.88$

Vedclass · Answer

(C) Given: Mole fraction of solvent $(X_{solvent})$ = $0.8$.Since it is an aqueous solution,the solvent is water $(H_2O)$,with molar mass $M_{solvent} = 18 \ g \ mol^{-1}$.Mole fraction of solute $(X_{solute})$ = $1 - X_{solvent} = 1 - 0.8 = 0.2$.Let the total number of moles be $n_{total} = 1$. Then $n_{solute} = 0.2 \ mol$ and $n_{solvent} = 0.8 \ mol$.Mass of solvent = $n_{solvent} 	imes M_{solvent} = 0.8 \ mol 	imes 18 \ g \ mol^{-1} = 14.4 \ g = 0.0144 \ kg$.Molality $(m)$ = $\frac{n_{solute}}{	ext{Mass of solvent in } kg} = \frac{0.2 \ mol}{0.0144 \ kg} = 13.88 \ mol \ kg^{-1}$.

The mole fraction of a solvent in an aqueous solution of a solute is $0.8$. The molality (in $mol \ kg^{-1}$) of the aqueous solution is:

Similar Questions

Write the formula for percentage by weight.

The density of a solution prepared by dissolving $120 \ g$ of urea (mol. mass $= 60 \ u$) in $1000 \ g$ of water is $1.15 \ g/mL$. The molarity of this solution is : ............ $M$

What will be the volume of a $12 \; M$ solution,if it is equivalent to $240 \; mL$ of an $18 \; M$ solution? (Express the answer in $litre$)

$3.0$ molal $NaOH$ solution has a density of $1.110 \ g/mL$. The molarity of the solution is :-

The density of a nitric acid solution is $1.5 \ g \ mL^{-1}$. Its weight percentage is $68 \%$. What is the approximate concentration (in $mol \ L^{-1}$) of nitric acid? $(N=14 \ u, O=16 \ u, H=1 \ u)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module