Search any question & find its solution
Question:
Answered & Verified by Expert
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?
Options:
Solution:
2357 Upvotes
Verified Answer
The correct answer is:
$\frac{M}{80}$
$\because$ 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 $=\frac{N}{40} \times \frac{1}{2}=\frac{M}{80}$
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 $=\frac{N}{40} \times \frac{1}{2}=\frac{M}{80}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.