2.20    A  5\% solution (by mass) of cane sugar in water has freezing point of  271 K.  Calculate the freezing point of  5\%  glucose in water if freezing point of pure water is 273.15 \; K.

Answers (1)

It is given that freezing point of pure water is 273.15 K.

So, elevation of freezing point = 273.15 - 271 = 2.15 K

5\% solution means 5 g solute in 95 g of water.

Moles of cane sugar :

                                       = \frac{5}{342} = 0.0146

Molality :  

                                       = \frac{0.0146}{0.095} = 0.1537

We also know that -              \Delta T_f = k_f \times m

or                                          k_f = \frac{\Delta T_f}{m} = 13.99\ K\ Kg\ mol^{-1}

Now we will use the above procedure for glucose.

5\% of glucose means 5 g of gluocse in 95 g of H2O.

Moles of glucose :

                               \frac{5}{180} = 0.0278

Thus molality :

                                     = \frac{.0278}{0.095} = 0.2926\ mol\ kg^{-1}

So,  we can find the elevation in freezing point:

                                                                               \Delta T_f = k_f \times m

                                                                                         = 13.99 \times 0.2926 = 4.09\ K

Thus freezing point of glucose solution is 273.15 - 4.09  = 269.06 K.

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