Density of a 2.05 M solution of H_2SO_4 in w | vedclass

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Question

(A) $1$. Normality $(N)$ is calculated as $N = M 	imes n	ext{-factor}$. For $H_2SO_4$,the $n	ext{-factor}$ is $2$. Thus,$N = 2.05 	imes 2 = 4.1 \

Vedclass · Accepted Answer

$4.1 \ N, 2.33 \ m$

Vedclass · Answer

(A) $1$. Normality $(N)$ is calculated as $N = M 	imes n	ext{-factor}$. For $H_2SO_4$,the $n	ext{-factor}$ is $2$. Thus,$N = 2.05 	imes 2 = 4.1 \ N$. $2$. Molality $(m)$ is calculated using the formula $m = \frac{1000 	imes M}{(1000 	imes d - M 	imes M_B)}$,where $M$ is molarity,$d$ is density,and $M_B$ is the molar mass of $H_2SO_4$ $(98 \ g/mol)$. $3$. Substituting the values: $m = \frac{1000 	imes 2.05}{(1000 	imes 1.08 - 2.05 	imes 98)} = \frac{2050}{(1080 - 200.9)} = \frac{2050}{879.1} \approx 2.33 \ m$.

Density of a $2.05 \ M$ solution of $H_2SO_4$ in water is $1.08 \ g/mL$. Find its Normality and molality.

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