The number of moles of solute present in the solutions of I, II and III is respectively I. \(500…
The number of moles of solute present in the solutions of I, II and III is respectively
I. \(500 \mathrm{~mL}\) of \(0.2 \mathrm{M} \mathrm{NaOH}\)
II. \(200 \mathrm{~mL}\) of \(0.1 \mathrm{~N} \mathrm{H}_2 \mathrm{SO}_4\)
III. \(6 \mathrm{~g}\) of urea in \(1 \mathrm{~kg}\) of water
\(0.1,0.01,0.1\)
\(0.1,0.02,0.1\)
\(0.2,0.01,0.1\)
\(0.1,0.01,0.2\)
Solution
I. Moles of solute \(\mathrm{NaOH}=\) molarity
\(=\frac{0.2 \times 500}{1000}=0.1 \mathrm{~mol}\)
II. Normality \(=n\)-factor \(\times\) molarity
\(\begin{aligned}
& \text { Molarity }\left(\mathrm{H}_2 \mathrm{SO}_4\right)=\frac{0.1}{2}=0.05 \mathrm{M} \\
& \text { moles of } \mathrm{H}_2 \mathrm{SO}_4=\frac{0.1 \times 200}{1000 \times 2}=0.01 \mathrm{M}
\end{aligned}\)
\(\begin{aligned}
\text { III. } \text { Moles }= & \frac{\text { Weight of urea }}{\text { Molecular weight of urea }} \\
& =\frac{6}{60}=0.1 \mathrm{~m}
\end{aligned}\)
Hence, option (1) is correct.