The number of moles of solute present in the solutions of I, II and III is respectively I. (500 ~mL ) of (0.2 M NaOH ) II. (200 ~mL ) of (0.1 ~N H ₂ SO ₄ ) III. (6 ~g ) of urea in (1 ~kg ) of water

Step-by-step solution

I. Moles of solute ( NaOH = ) molarity (= 0.2 500 1000 =0.1 ~mol ) II. Normality (=n )-factor ( ) molarity ( aligned & Molarity ( H ₂ SO ₄ )= 0.1 2 =0.05 M & moles of H ₂ SO ₄= 0.1 200 1000 2 =0.01 M aligned ) ( aligned III. Moles = & Weight of urea Molecular weight of urea & = 6 60 =0.1 ~m aligned ) Hence, option (1) is correct.