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  • Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water,it depresses the freezing point by <img alt="0.2^{\circ} \mathrm{C}" src="https://entrancecorner.oncodecogs

Solute A associates in water. When 0.7 g of solute A is dissolved in 42.0 g of water,it depresses the freezing point by 0.2^{\circ} \mathrm{C}. The percentage association of solute A in water is:

[Given: Molar mass of \mathrm{A}=93 \mathrm{~g} \mathrm{~mol}^{-1}. Molal depression constant of water is \left.1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} .\right]

Option: 1

50%


Option: 2

60%


Option: 3

70%


Option: 4

80%


Answers (1)

best_answer

Solute A associates in water.

So, \\\mathrm{2A\rightleftharpoons A_{2}}\\1\; \; \; \; \; \; \; \; \; 0

  \\\mathrm{1-\alpha \; \; \; \; \; \; \; \frac{\alpha }{2}}, \mathrm{\alpha =\text{% association of solute A in water}}

The van't Hoff factor can be calculated as 

\begin{aligned} &\mathrm{i=1-\alpha+\frac{\alpha}{2} }\\ &\mathrm{i=1-\frac{\alpha}{2}}\; \; \; \quad \rightarrow (1) \end{aligned}

Now given, 

\mathrm{\Delta T_{f}=0.2^{\circ} C=0.2 \mathrm{~K}, \quad K_{f}=1.86 \mathrm{~K}\; \mathrm{Kg}\; \; \mathrm{mol}^{-1}}

\begin{aligned} &\mathrm{n_{A}=\frac{W_ A}{M_{A}}=\frac{0.7}{93}} \\ &\mathrm{W_{B}=42.0 \mathrm{~g}=\frac{42.0}{1000} \mathrm{~kg}} \end{aligned}

We know, 

\mathrm{\Delta T_{f}=i\; \mathrm{kf} \; m }

\Rightarrow\mathrm{ 0.2=i \times(1.86) \times \frac{0.7}{93} \times \frac{1000}{42.0}}

  \Rightarrow \mathrm{i=0.60} \quad \rightarrow (2)

Putting value of i in (1) , we have

\mathrm{0.60=1-\frac{\alpha }{2}\Rightarrow \alpha =0.80}

\therefore %\; \alpha =80%

Hence, the correct answer is Option (4)

Posted by

avinash.dongre

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