The molarity of a 70 % (mass/mass) aqueous so | vedclass

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(C) Let the mass of the solution be $100 \ g$. Since the solution is $70 \%$ (mass/mass),the mass of the solute $(X)$ is $70 \ g$. The volume of the s

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Let the mass of the solution be $100 \ g$. Since the solution is $70 \%$ (mass/mass),the mass of the solute $(X)$ is $70 \ g$. The volume of the solution is calculated as: $V = \frac{	ext{mass}}{	ext{density}} = \frac{100 \ g}{1.25 \ g \ mL^{-1}} = 80 \ mL$. The number of moles of the acid $(X)$ is: $n = \frac{	ext{mass}}{	ext{molar mass}} = \frac{70 \ g}{70 \ g \ mol^{-1}} = 1 \ mol$. Molarity $(M)$ is defined as: $M = \frac{n 	imes 1000}{V 	ext{ (in } mL)} = \frac{1 	imes 1000}{80} = 12.5 \ M$. Rounding to the nearest integer,we get $13 \ M$.

The molarity of a $70 \%$ (mass/mass) aqueous solution of a monobasic acid $(X)$ is $........... \ M$ (Nearest integer). Given: Density of aqueous solution of $(X)$ is $1.25 \ g \ mL^{-1}$,Molar mass of the acid is $70 \ g \ mol^{-1}$.

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