Isotonic solutions are those which have same osmotic pressure (П = 𝑖𝐶𝑅𝑇). But here we have different concentration of the solutions and also they have different van’t Hoff factors (i). So the solutions for which the product of i and c will be same, will be isotonic. a. For, 0.001 M ⇒, 𝑖 = 3. So 𝑖 × 𝐶 = 3 × 0.001 = 0.003 For, 0.001 M 𝐴𝑙2(𝑆𝑂4)3 ,𝑖 = 5. So 𝑖 × 𝐶 = 5 × 0.001 = 0.005 b. For, 0.01 M 𝐵𝑎𝐶𝑙2 , 𝑖 = 3. So 𝑖 × 𝐶 = 3 × 0.01 = 0.03 For, 0.001 M 𝐶𝑎𝐶𝑙2 , 𝑖 = 3. So 𝑖 × 𝐶 = 3 × 0.001 = 0.003 c. For, 0.01 M ⇒, 𝑖 = 3. So 𝑖 × 𝐶 = 3 × 0.01 = 0.03 For, 0.015 M 𝑁𝑎𝐶𝑙, 𝑖 = 2. So 𝑖 × 𝐶 = 2 × 0.015 = 0.03 Thus 0.01 M 𝐵𝑎𝐶𝑙2 and 0.015 M 𝑁𝑎𝐶𝑙 are isotonic in nature. d. For, 0.001 M 𝐴𝑙2(𝑆𝑂4)3, 𝑖 = 5. So 𝑖 × 𝐶 = 5 × 0.001 = 0.005 For, 0.01 M 𝑎𝐶𝑙2,𝑖 = 3. So 𝑖 × 𝐶 = 3 × 0.01 = 0.03