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To neutralise $20 \mathrm{~m} /$ of $M / 10$ sodium hydroxide, the volume of $M / 20$ hydrochloric acid required is
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$40 \mathrm{ml}$
$\mathrm{NaOH} \quad \mathrm{HCl}$
$N_1 V_1=N_2 V_2 ; \quad 20 \times \frac{1}{10}=\frac{1}{20} \times V ; V=40 \mathrm{ml}$
$N_1 V_1=N_2 V_2 ; \quad 20 \times \frac{1}{10}=\frac{1}{20} \times V ; V=40 \mathrm{ml}$
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