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If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution, what is the change in freezing point of water$(\Delta T_f)$,when 0.02mol of sodium sulphate is dissolved in 1kg of water,is $(K_f=1.86Kkgmol^{-1})$?

$\begin{array}{1 1}(a)\;0.0372K\\(b)\;0.112K\\(c)\;0.0744K\\(d)\;0.0186K\end{array}$

1 Answer

Answer: 0.1120$^{\circ}$K
Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.
Calculating $i$ for Sodium Sulfate $\rightarrow Na_2SO_4^{2-} (s) \rightarrow 2Na^{+} + SO_4^{2-} \rightarrow i = 3$
In this case $i = 3, \; K_f = 1.86$ and $m = 0.02 \rightarrow \Delta T_f = 3 \times 1.86 \times 0.02 = 0.112$
answered Feb 13, 2014 by sreemathi.v
edited Jul 16, 2014 by balaji.thirumalai

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