10.0 mL of 0.0500 M K2Cr2O7 is treated with excess KI in acidic solution (Cr2O7^2- + 6I- + 14H+ -> 2Cr^3+ + 3I2 + 7H2O). The liberated iodine is titrated with 0.100 M Na2S2O3. The volume of thiosulphate needed is:

JEE Main levelSingle correct optionDifficulty 2 of 3rx184

10.0 mL of 0.0500 M $K X_{2} Cr X_{2} O X_{7}$ is treated with excess KI in acidic solution ( $Cr X_{2} O X_{7} X^{2 -} + 6 I X^{-} + 14 H X^{+} 2 Cr X^{3 +} + 3 I X_{2} + 7 H X_{2} O$ ). The liberated iodine is titrated with 0.100 M $Na X_{2} S X_{2} O X_{3}$ . The volume of thiosulphate needed is:

mmol dichromate = 0.500, so mmol

I X_{2}

0.500 \times 3 = 1.50

, and mmol thiosulphate =

1.50 \times 2 = 3.00

. Volume = 3.00/0.100 = 30.0 mL.