25.0 mL of an acidified SnCl2 solution is titrated with 0.0200 M K2Cr2O7 and requires 18.4 mL. In the reaction Sn^2+ is oxidised to Sn^4+ and Cr2O7^2- is reduced to Cr^3+. Taking Sn = 119 and Cl = 35.5, find the concentration of SnCl2 in the solution in g/L (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3pb110

25.0 mL of an acidified $SnCl X_{2}$ solution is titrated with 0.0200 M $K X_{2} Cr X_{2} O X_{7}$ and requires 18.4 mL. In the reaction $Sn X^{2 +}$ is oxidised to $Sn X^{4 +}$ and $Cr X_{2} O X_{7} X^{2 -}$ is reduced to $Cr X^{3 +}$ . Taking Sn = 119 and Cl = 35.5, find the concentration of $SnCl X_{2}$ in the solution in g/L (two decimals).

Numerical answer type, answer in g/L, tolerance 0.02.

Cr X_{2} O X_{7} X^{2 -}

takes 6 electrons and each

Sn X^{2 +}

gives 2, so

Cr X_{2} O X_{7} X^{2 -} : Sn X^{2 +} = 1 : 3

n (Cr X_{2} O X_{7} X^{2 -}) = 0.0200 \times 18.4 = 0.368

mmol, so

n (Sn X^{2 +}) = 1.104

mmol in 25.0 mL, which is 0.04416 M. Molar mass of

SnCl X_{2}

= 119 + 71 = 190

g/mol, so the concentration

= 0.04416 \times 190 = 8.3904

g/L, i.e. 8.39 g/L.