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(D) Step $1$: Calculate the normality of $Na_2CO_3$ solution.Normality of $HCl$ = $\frac{109.5 \ g/L}{36.5 \ g/eq} = 3 \ N$.Using $N_1V_1 = N_2V_2$ fo

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$470$

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(D) Step $1$: Calculate the normality of $Na_2CO_3$ solution.Normality of $HCl$ = $\frac{109.5 \ g/L}{36.5 \ g/eq} = 3 \ N$.Using $N_1V_1 = N_2V_2$ for $Na_2CO_3$ and $HCl$:$N_{Na_2CO_3} 	imes 25 \ mL = 3 \ N 	imes 32.9 \ mL$.$N_{Na_2CO_3} = \frac{3 	imes 32.9}{25} = 3.948 \ N$.Step $2$: Calculate the volume of $Na_2CO_3$ solution corresponding to $125 \ g$.Volume = $\frac{	ext{mass}}{	ext{density}} = \frac{125 \ g}{1.25 \ g/mL} = 100 \ mL$.Step $3$: Calculate the volume of $0.84 \ N$ $H_2SO_4$ required to neutralize $100 \ mL$ of $3.948 \ N$ $Na_2CO_3$ solution.$N_3V_3 = N_4V_4$$3.948 \ N 	imes 100 \ mL = 0.84 \ N 	imes V_{H_2SO_4}$.$V_{H_2SO_4} = \frac{394.8}{0.84} = 470 \ mL$.

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