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A sample of water is found to contain $5.85 \%$ $\left(\frac{w}{w}\right)$ of $A B$ (molecular mass 58.5 ) and $9.50 \%$ $\left(\frac{w}{w}\right) X Y_2$ (molecular mass 95). Assuming $80 \%$ ionisation of $A B$ and $60 \%$ ionisation of $X_2$, the freezing point of water sample is [Given, $K_f$ for water $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$,
Freezing point of pure water is $273 \mathrm{~K}$ and $A$, $B$ and $Y$ are monovalent ions.]
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Freezing point of pure water is $273 \mathrm{~K}$ and $A$, $B$ and $Y$ are monovalent ions.]
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Verified Answer
The correct answer is:
$264.25 \mathrm{~K}$
$A B \quad X Y_2$
$5.85 \%$
$9.5 \%$
$\begin{array}{rlrl}m & =\frac{5.85}{58.5} \times \frac{1000}{84.65} \quad m & =\frac{9.5}{95} \times \frac{1000}{84.65} \\ & =1.181 & & =1.181\end{array}$
$\alpha=0.8$ (for $80 \%$ ionisation) $\alpha=0.6$
(for $60 \%$ ionisation)
$\begin{array}{rlrl}i & =1+(2-1) 0.8 & i & =1+(3-1) 0.6 \\ & =1.8 & & =2.2\end{array}$
Now, $\Delta T_f=K_f\left[i_1 m_1+i_2 m_2\right]$
$\begin{aligned} & =1.86[1.8(1.181)+2.2(1.181)] \\ & =1.86[4.72]=8.78\end{aligned}$
$5.85 \%$
$9.5 \%$
$\begin{array}{rlrl}m & =\frac{5.85}{58.5} \times \frac{1000}{84.65} \quad m & =\frac{9.5}{95} \times \frac{1000}{84.65} \\ & =1.181 & & =1.181\end{array}$
$\alpha=0.8$ (for $80 \%$ ionisation) $\alpha=0.6$
(for $60 \%$ ionisation)
$\begin{array}{rlrl}i & =1+(2-1) 0.8 & i & =1+(3-1) 0.6 \\ & =1.8 & & =2.2\end{array}$
Now, $\Delta T_f=K_f\left[i_1 m_1+i_2 m_2\right]$
$\begin{aligned} & =1.86[1.8(1.181)+2.2(1.181)] \\ & =1.86[4.72]=8.78\end{aligned}$
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