$25.0 mL$ of $0.0500 M$ $K X_{2} Cr X_{2} O X_{7}$ is treated with excess $KI$ in acid and the liberated iodine requires $36.4 mL$ of sodium thiosulphate solution. The molarity of the thiosulphate is

A0.412 M
B0.206 M
C0.103 M
D0.309 M

Answer: B

mol Cr2O7(2-) = 0.0250 x 0.0500 = 1.25e-3; I2 = 3 x that = 3.75e-3; thiosulphate = 2 x I2 = 7.50e-03 mol (6 equivalents per dichromate). Molarity = 7.50e-03/0.0364 L = 0.206 M.

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Answer: B

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$25.0 mL$ of $0.0500 M$ $K X_{2} Cr X_{2} O X_{7}$ is treated with excess $KI$ in acid and the liberated iodine requires $36.4 mL$ of sodium thiosulphate solution. The molarity of the thiosulphate is