Q21 / 160 · AP EAMCET 2024 (21 May Shift 1) · Chemistry
100 mL of $\frac{M}{10} \mathrm{Ca}\left(\mathrm{NO}_3\right)_2$ and 200 mL of $\frac{M}{10} \mathrm{KNO}_3$ solutions are mixed. What is the normality of resulted solution with respect to $\mathrm{NO}_3^{-}$?
- A0.1 N
- B0.2 N
- Correct answerC0.13 N
- D0.066 N
Solution
We know that
Normality $=$ Molarity $\times \mathrm{Z}$ factor $\mathrm{N}_1=\frac{1}{10} \times 2=\frac{1}{5}$ $\mathrm{N}_2=\frac{1}{10} \times 1=\frac{1}{10}$ $\therefore$ no. of mole $\left(\mathrm{n}_1\right)=\frac{1}{5} \times 100=20$ mole no. of mole $\left(n_2\right)=\frac{1}{10} \times 200=20$ mole Total mole $\left(\mathrm{n}_1+\mathrm{n}_2\right)=40 \mathrm{~mole}$ Concentration/Normality of solution $=\frac{n_1+n_2}{v_1+v_2}$ $=\frac{40}{300}=0.13 \mathrm{~N}$
Normality $=$ Molarity $\times \mathrm{Z}$ factor $\mathrm{N}_1=\frac{1}{10} \times 2=\frac{1}{5}$ $\mathrm{N}_2=\frac{1}{10} \times 1=\frac{1}{10}$ $\therefore$ no. of mole $\left(\mathrm{n}_1\right)=\frac{1}{5} \times 100=20$ mole no. of mole $\left(n_2\right)=\frac{1}{10} \times 200=20$ mole Total mole $\left(\mathrm{n}_1+\mathrm{n}_2\right)=40 \mathrm{~mole}$ Concentration/Normality of solution $=\frac{n_1+n_2}{v_1+v_2}$ $=\frac{40}{300}=0.13 \mathrm{~N}$
Hence, the correct answer is option: 'C'.