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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 $g$)?

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