A solution is prepared by adding 124 g of et | vedclass

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

Question

(D) The molality $(m)$ of a solution is defined as the number of moles of solute per kilogram of solvent. Given: Mass of ethylene glycol $= 124 \ g$.

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(D) The molality $(m)$ of a solution is defined as the number of moles of solute per kilogram of solvent. Given: Mass of ethylene glycol $= 124 \ g$. Molar mass of ethylene glycol $= 62 \ g \ mol^{-1}$. Moles of ethylene glycol $(n)$ $= \frac{124 \ g}{62 \ g \ mol^{-1}} = 2 \ mol$. Molality $(m)$ $= 10 \ mol \ kg^{-1}$. Formula: $m = \frac{n_{	ext{solute}}}{W_{	ext{solvent (in kg)}}}$. Substituting the values: $10 = \frac{2}{W_{	ext{solvent (in kg)}}}$. $W_{	ext{solvent (in kg)}} = \frac{2}{10} = 0.2 \ kg$. Since $1 \ kg = 1000 \ g$,$W_{	ext{solvent}} = 0.2 	imes 1000 = 200 \ g$. Therefore,$x = 200$.

$A$ solution is prepared by adding $124 \ g$ of ethylene glycol (molar mass $= 62 \ g \ mol^{-1}$) to $x \ g$ of water to get a $10 \ m$ solution. What is the value of $x$ (in $g$)?

Similar Questions

Calculate the molarity of a solution prepared by dissolving $7.1 \ g$ of sodium sulfate $(Na_2SO_4)$ in water to make $100 \ mL$ of solution. (in $M$)

By dissolving $0.35 \, \text{mole}$ of sodium chloride in water,$1.30 \, \text{L}$ of salt solution is obtained. The molarity of the resulting solution should be reported as $..... \, \text{M}$.

What is the mole fraction of water in a spirit sample containing $92\%$ ethyl alcohol by mass?

$A$ solution is obtained by mixing $300 \, g$ of $25 \%$ solution and $400 \, g$ of $40 \%$ solution by mass. Calculate the mass percentage of the resulting solution. (in $\%$)

An aqueous solution of glucose is $10\%$ in strength. The volume in which $1 \, \text{g-mole}$ of it is dissolved will be ........... $L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module