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  • (a) A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water. The freezing point of pure water is 273·15 K. (b) Why is osmotic pressure of 1

(a) A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% solution (by mass) of glucose in water. The freezing point of pure water is 273·15 K.

(b) Why is osmotic pressure of 1 M KCl higher than 1 M urea solution ?

(c) What type of liquids form ideal solutions ?

 

 

 

 
 
 
 
 

Answers (1)

(a) 

\begin{aligned} \Delta T_{f} &=T_{f}^{o}-I_{f} \\ &=273.15-271 \\ &=2.15 \mathrm{K} \end{aligned}

5% solution means = (100-5) g

                                = 95g of water

No. of moles of cane sugar  =\frac{5}{342}mol

No. of moles of glucose =\frac{5}{180}mol

Molality of cane sugar =\frac{5}{342}\times \frac{1}{0.095}

                                    =0.1537\: kg^{-1}mol^{-1}

Molality of glucose =\frac{5}{180}\times \frac{1}{0.095}=0.2926\: molkg^{-1}mol^{-1}

For cane sugar,

\begin{aligned} & \Delta T_{f}=K_{f} \times m \\ K_{f} &=\frac{\Delta T_{f}}{m}=\frac{2.15}{0.1537} \\ &=13.99 \mathrm{K} \mathrm{kg} / \mathrm{mol} \end{aligned}

For glucose,

\begin{aligned} \Delta T_{f} &=K_{f} \times m \\ \Delta T_{f} &=13.99 \times 0.2926 \\ &=4.09 \mathrm{K} \end{aligned}

freezing point of glucose,

\begin{aligned} &T_{f}=T_{f}^{\circ}-\Delta T_{f}\\ &=273.15-4.09\\ &=269.06 K \end{aligned}

(b) Osmotic pressure of 1M KCl is higher than 1M urea solution as KCl ionises into  K^+ and  Cl^- and osmotic pressure being colligative property depends on number of particles and urea does not ionise at all.

(c) A solution of two liquids behave as an ideal solution if unlike attractions in them are equal to like attractions.

 

 

Posted by

Sumit Saini

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