Calculate the mole fraction of ethylene glyco | vedclass

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

Question

(A) Assume that we have $100 \ g$ of solution.Solution will contain $20 \ g$ of ethylene glycol and $80 \ g$ of water.Molar mass of $C_{2}H_{6}O_{2} =

Vedclass · Accepted Answer

$0.068$

Vedclass · Answer

(A) Assume that we have $100 \ g$ of solution.Solution will contain $20 \ g$ of ethylene glycol and $80 \ g$ of water.Molar mass of $C_{2}H_{6}O_{2} = 12 	imes 2 + 1 	imes 6 + 16 	imes 2 = 62 \ g \ mol^{-1}$.Moles of $C_{2}H_{6}O_{2} = \frac{20 \ g}{62 \ g \ mol^{-1}} = 0.322 \ mol$.Moles of water = $\frac{80 \ g}{18 \ g \ mol^{-1}} = 4.444 \ mol$.Mole fraction of ethylene glycol $(x_{glycol})$ = $\frac{	ext{moles of } C_{2}H_{6}O_{2}}{	ext{moles of } C_{2}H_{6}O_{2} + 	ext{moles of } H_{2}O}$.$x_{glycol} = \frac{0.322}{0.322 + 4.444} = \frac{0.322}{4.766} = 0.068$.

Calculate the mole fraction of ethylene glycol $(C_{2}H_{6}O_{2})$ in a solution containing $20\%$ of $C_{2}H_{6}O_{2}$ by mass.

Similar Questions

The mole fraction of the solute in a $1 \ m$ (molal) aqueous 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? $(N=14 \ u, O=16 \ u, H=1 \ u)$

$25 \ g$ of a solute of molar mass $250 \ g \ mol^{-1}$ is dissolved in $100 \ mL$ of water to obtain a solution whose density is $1.25 \ g \ mL^{-1}$. The molarity and molality of the solution are respectively

What is the mole fraction of the solute in a $1.00 \, m$ aqueous solution?

The density of a $1.0 \ M$ aqueous solution of glucose is $1.1 \ g/mL$. What is the molality of the solution in $m$ (in $m$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module