Q7 / 160 · AP EAMCET 2018 (22 Apr Shift 2) · Chemistry

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?

  • A
    $\frac{M}{40}$
  • B
    $\frac{M}{20}$
  • Correct answer
    C
    $\frac{M}{80}$
  • D
    $\frac{M}{30}$

Solution

$\because$ Initial volume $\left(V_1\right)=50 \mathrm{~mL}$ Initial normality $\left(N_1\right)=0.1 \mathrm{~N}$ Final volume $\left(V_2\right)=50+150=200 \mathrm{~mL}$ Final normality $\left(N_2\right)=$ To find and, $N_1 \times V_1=N_2 \times V_2 \quad 0.1 \times 50=(=N) \times 200$ $\therefore \quad N_2=\frac{(0.1 \times 50) N}{200}=\frac{N}{40}$ 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 $=\frac{N}{40} \times \frac{1}{2}=\frac{M}{80}$

Hence, the correct answer is option: 'C'.

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