The weight of H_2SO_4 in 150 mL of N = 1/7 s | vedclass

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(B) The formula for normality is $N = \frac{	ext{Weight in gram}}{	ext{Equivalent weight} 	imes 	ext{Volume in Liters}}$.Equivalent weight of $H_2

Vedclass · Accepted Answer

$1.05$

Vedclass · Answer

(B) The formula for normality is $N = \frac{	ext{Weight in gram}}{	ext{Equivalent weight} 	imes 	ext{Volume in Liters}}$.Equivalent weight of $H_2SO_4 = \frac{	ext{Molar mass}}{	ext{Valency factor}} = \frac{98}{2} = 49$.Weight in gram = $N 	imes 	ext{Equivalent weight} 	imes 	ext{Volume in Liters}$.Weight = $\frac{1}{7} 	imes 49 	imes \frac{150}{1000}$.Weight = $7 	imes 0.15 = 1.05 \, g$.

The weight of $H_2SO_4$ in $150 \ mL$ of $N = 1/7$ solution is ........ $g$.

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