Q21 / 160 · AP EAMCET 2019 (20 Apr Shift 1) · Chemistry

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

  • Correct answer
    A
    \(0.1,0.01,0.1\)
  • B
    \(0.1,0.02,0.1\)
  • C
    \(0.2,0.01,0.1\)
  • D
    \(0.1,0.01,0.2\)

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.

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

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