To $50 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{Na}_2 \mathrm{CO}_3$ solution $150 \mathrm{~mL}$ of water is added. What is the molarity of resultant solution?

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Aicharya · Accepted Answer

$\because$ Initial volume $\left(V_1ight)=50 \mathrm{~mL}$
Initial normality $\left(N_1ight)=0.1 \mathrm{~N}$
Final volume $\left(V_2ight)=50+150=200 \mathrm{~mL}$
Final normality $\left(N_2ight)=$ To find
and, $N_1 \times V_1=N_2 \times V_2$
$0.1 \times 50=(N_2) \times 200$
$\therefore \quad N_2=\frac{(0.1 \times 50) N}{200}=\frac{N}{40}$
Also, $\mathrm{Z}$ for $\mathrm{Na}_2 \mathrm{CO}_3=2$
(where, $Z=$ total positive charge in $\mathrm{Na}_2 \mathrm{CO}_3$ )
and $N$ (normality) $=M \times Z$ (where, $M$ is molarity).
$\therefore$ Molarity of resultant solution $=\frac{N}{40} \times \frac{1}{2}=\frac{M}{80}$ .

To $50 \mathrm{~mL}$ of $0.1 \mathrm{~N} \mathrm{Na}_2 \mathrm{CO}_3$ solution $150 \mathrm{~mL}$ of water…

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