KCET · Chemistry · Solutions
Solute ' \(X^{\prime}\) dimerises in water to the extent of \(80 \% .2 .5 \mathrm{~g}\) of ' \(X^{\prime}\) in \(100 \mathrm{~g}\) of water increases the boiling point by \(0.3^{\circ} \mathrm{C}\). The molar mass of \(' X^{\prime}\) is \(\left[K_{b}=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}\right]\)
- A 13
- B 52
- C 65
- D 26
Answer & Solution
Correct Answer
(D) 26
Step-by-step Solution
Detailed explanation
Here, \(2.5 \mathrm{~g} X\) in \(100 \mathrm{~g}\) water is increasing boiling point by \(0.3^{\circ} \mathrm{C}\) and \(X\) is getting dimerise to the extent of \(80 \%\).
Molar mass of \(X=\) ?
\(K_{b}=0.52\)
Here,
\(i=\frac{1-\alpha+\frac{\alpha}{2}}{1}=1-0.8+\frac{0.8}{2}\) (Due to \(80 \%\) dimerisation)
\(i=0.6\)
Now, \(M=K_{b}\left(\frac{w_{B} \times 1000}{w_{A} \times \Delta T_{b}}\right) \times i\)
\(\begin{aligned}
&=0.52\left(\frac{2.5 \times 1000}{100 \times 0.3}\right)=\frac{52}{100} \times \frac{25}{3} \times 10 \times 0.6 \\
&=26
\end{aligned}\)
Molar mass of \(X=\) ?
\(K_{b}=0.52\)
Here,
\(i=\frac{1-\alpha+\frac{\alpha}{2}}{1}=1-0.8+\frac{0.8}{2}\) (Due to \(80 \%\) dimerisation)
\(i=0.6\)
Now, \(M=K_{b}\left(\frac{w_{B} \times 1000}{w_{A} \times \Delta T_{b}}\right) \times i\)
\(\begin{aligned}
&=0.52\left(\frac{2.5 \times 1000}{100 \times 0.3}\right)=\frac{52}{100} \times \frac{25}{3} \times 10 \times 0.6 \\
&=26
\end{aligned}\)
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