When 1 L of N/5 HCl aqueous solution in a b | vedclass

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Question

(C) Initial amount of $HCl$ in $1 \ L$ of $N/5 \ HCl$ solution: $Normality (N) = \frac{	ext{Weight of solute (g)}}{	ext{Equivalent weight} 	imes

Vedclass · Accepted Answer

$N/2.5$

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(C) Initial amount of $HCl$ in $1 \ L$ of $N/5 \ HCl$ solution: $Normality (N) = \frac{	ext{Weight of solute (g)}}{	ext{Equivalent weight} 	imes 	ext{Volume of solution (L)}}$ For $HCl$,Equivalent weight = Molar mass = $36.5 \ g/mol$. Initial weight of $HCl = N 	imes 	ext{Equivalent weight} 	imes 	ext{Volume (L)} = (1/5) 	imes 36.5 	imes 1 = 7.3 \ g$. After boiling,$3.65 \ g$ of $HCl$ is lost. Remaining weight of $HCl = 7.3 \ g - 3.65 \ g = 3.65 \ g$. The volume of the resulting solution is $250 \ mL = 0.25 \ L$. New Normality $(N') = \frac{	ext{Remaining weight of } HCl}{	ext{Equivalent weight} 	imes 	ext{Volume (L)}} = \frac{3.65}{36.5 	imes 0.25} = \frac{1}{10 	imes 0.25} = \frac{1}{2.5} = N/2.5$.

When $1 \ L$ of $N/5 \ HCl$ aqueous solution in a beaker is boiled,the volume of the resulting solution becomes $250 \ mL$. During this process,$3.65 \ g$ of $HCl$ is lost. What will be the concentration of the resulting solution? $(HCl = 36.5 \ g/mol)$

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