A solution contains 25% {H_2}O,25% {C_2}{H_5} | vedclass

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

Question

(C) Let the total mass of the solution be $100 \ g$. Mass of ${H_2}O = 25 \ g$,Mass of ${C_2}{H_5}OH = 25 \ g$,and Mass of ${CH_3}COOH = 50 \ g$. Mole

Vedclass · Accepted Answer

$0.502$

Vedclass · Answer

(C) Let the total mass of the solution be $100 \ g$. Mass of ${H_2}O = 25 \ g$,Mass of ${C_2}{H_5}OH = 25 \ g$,and Mass of ${CH_3}COOH = 50 \ g$. Moles of ${H_2}O = \frac{25 \ g}{18 \ g/mol} = 1.389 \ mol$. Moles of ${C_2}{H_5}OH = \frac{25 \ g}{46 \ g/mol} = 0.543 \ mol$. Moles of ${CH_3}COOH = \frac{50 \ g}{60 \ g/mol} = 0.833 \ mol$. Total moles $= 1.389 + 0.543 + 0.833 = 2.765 \ mol$. Mole fraction of ${H_2}O = \frac{	ext{moles of } {H_2}O}{	ext{total moles}} = \frac{1.389}{2.765} \approx 0.502$.

$A$ solution contains $25\%$ ${H_2}O$,$25\%$ ${C_2}{H_5}OH$ and $50\%$ ${CH_3}COOH$ by mass. The mole fraction of ${H_2}O$ would be

Similar Questions

$3.0 \ g$ ethanoic acid in $50 \ g$ benzene has . . . . . . molality. (Atomic weights: $H=1, C=12, O=16$).

The molarity of pure water at $4\,^{\circ}C$ is ............... $M$.

The number of moles of a solute in its solution is $20$ and the total number of moles is $80$. The mole fraction of the solute is:

The density of $3 \ M$ solution of $NaCl$ is $1.25 \ g \ mL^{-1}$. Calculate the molality of the solution.

The volume of water which must be added to $40 \ cc$ of $0.25 \ M$ oxalic acid solution to make it exactly decimolar $(0.1 \ M)$ is $....... \ cc$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module