Q

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

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

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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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