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To $500cm^3$ of water $3.0\times 10^{-3}$kg of acetic acid is added .If 2.3% of acetic acid is dissociated.What will be the depression of freezing point?$K_f$ and density of water are $1.86$K kg/mol and $0.997g/cm^3$ respectively.

$\begin{array}{1 1}0.23K\\0.43K\\0.63K\\0.13K\end{array} $

1 Answer

Mass of acetic acid =$3\times 10^{-3}kg=3g$
Amount of acetic acid ,
$n_2=\large\frac{3g}{60g mol^{-1}}$$=0.05$ mol
(Molar mass of acetic acid =60)
Volume of water,$V=500cm^3$
Mass of water,$m_1=V_{\rho}=(500cm^3)(0.997gcm^{-3})$
$\Rightarrow 498.5g=0.4985kg$
Molality of acetic acid,m=$\large\frac{n_2}{m_1}$
$\Rightarrow \large\frac{0.05mol}{0.4985kg}=$$0.1003mol kg^{-1}$
Since 23% acetic acid is dissociated,its degree of dissociation will be 0.23
Hence in solution we have
$CH_3COOH \quad \leftrightharpoons\quad CH_3COO^-+H^+$
$(1-\alpha)\qquad\qquad\qquad \alpha\qquad\quad\;\; \alpha$
The total number of species in solution
As,$-\Delta T_f=iK_f m$
$\Delta T=(1+\alpha)K_f m$
$\Rightarrow (0.1003mol kg^{-1})(1+0.23)(1.86Kkg mol^{-1})$
$\Rightarrow 0.23K$
answered Aug 6, 2014 by sreemathi.v

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