IIT-JEE 2011 · Paper 2 · Chemistry Q5

ChemistrySingle correct+3 / -1Question 5 of 60

The freezing point (in °C) of a solution containing 0.1 g of $\ce{K3[Fe(CN)6]}$ (Mol. Wt. 329) in 100 g of water ( $K_f=1.86$ K kg mol $^{-1}$ ) is

-2.3\times10^{-2}

-5.7\times10^{-2}

-5.7\times10^{-3}

-1.2\times10^{-2}

Answer and worked solution

Official answer: A

Van 't Hoff factor. $\ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^{3-}}$ , so $i=4$ .

Molality. $m=\dfrac{0.1/329}{0.1}=3.04\times10^{-3}$ mol kg⁻¹.

Depression. $\Delta T_f=iK_fm=4\times1.86\times3.04\times10^{-3}=2.3\times10^{-2}$ K, so the freezing point is $-2.3\times10^{-2}$ °C.

(A)

Topics:Colligative properties van't Hoff factor

IIT-JEE 2011 · Paper 2, Chemistry question 5