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KCET · Chemistry · Solutions

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.]

  1. A \(264.25 \mathrm{~K}\)
  2. B \(265.56 \mathrm{~K}\)
  3. C \(280.44 \mathrm{~K}\)
  4. D \(281.75 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(264.25 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\(A B \quad X Y_2\)
\(5.85 \%\)
\(9.5 \%\)
\(\begin{array}{rlrl}m & =\frac{5.85}{58.5} \times \frac{1000}{84.65} \quad m & =\frac{9.5}{95} \times \frac{1000}{84.65} \\ & =1.181 & & =1.181\end{array}\)
\(\alpha=0.8\) (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, \(\Delta T_f=K_f\left[i_1 m_1+i_2 m_2\right]\)
\(\begin{aligned} & =1.86[1.8(1.181)+2.2(1.181)] \\ & =1.86[4.72]=8.78\end{aligned}\)