$(a)\;350ml\qquad(b)\;470ml\qquad(c)\;225ml\qquad(d)\;330ml$

$N_{HCl}=\large\frac{109.5}{36.5\times 1}$$=3$

$Na_2CO_3$ is completely neutralized by HCl

Meq of $Na_2CO_3$=Meq of $HCl$

$N\times 25=32.9\times 3$

$N_{Na_2CO_3}=3.948$

Weight of $Na_2CO_3$ solution=125g

Volume of $Na_2CO_3$ solution =$\large\frac{125}{1.25}$$=100ml$

$\therefore$ Meq of $H_2SO_4$=Meq of $Na_2CO_3$

$0.84\times V=100\times 3.948$

Volume of $H_2SO_4$ required =470ml

Hence (b) is the correct answer.

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