Question

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

Answer 1

Hint: Depression in freezing point is one of the colligative properties of solutions. Here, there is the comparison among the cane sugar and glucose for the calculations of depression in freezing point.

Complete step by step answer:
Let’s start with understanding the concept of depression in freezing point (Cryoscopy).
Freezing point-
The temperature at which the vapour pressure of liquid is equal to the vapour pressure of solid formed is called the freezing point.
Vapour pressure of liquid = Vapour pressure of solid

Depression in freezing point of the solution-
On adding non-volatile solute to pure solvent, the freezing point of the solution will decrease than that of pure solvent and this decrease is called depression in freezing point. The liquid solution has a lower freezing point than pure solvent because the chemical potential of the solvent in the mixture is lower than that of pure solvent.
Example:
In general, salt added in smaller amounts can be considered as solute and water which is originally present in larger quantities can be considered as solvent. Thus, the freezing point of pure solvent (water) would be higher than that of the resulting solution (salt-water mixture). Hence, freezing point depression took place when salt was added to water.
It is stated as,
Depression in freezing point = (Freezing point of pure solvent) - (Freezing point of solution)
i.e. $\Delta {{T}_{f}}$ = (Freezing point of pure solvent) - (Freezing point of solution)

For dilute solution-
$\Delta {{T}_{f}} = {{K}_{f}}\times m\times i$
where,
${{K}_{f}}$ = Molal depression constant (Cryoscopic constant)
m = Molality
i = Van't Hoff’s factor

Illustration-
The depression in freezing point for cane sugar (sucrose);
$\Delta {{T}_{f}} = 273.15 - 271 = 2.15K$
As, 5% of solute is present in the solution thus, 5 gm of solute is present in 95 gm of solvent (for both sucrose and glucose).
The molar masses of glucose and sucrose are 180 gm/mol and 342 gm/mol respectively.
Thus,
Number of moles of glucose = $\dfrac{5}{180}=0.028 moles$
Number of moles of sucrose = $\dfrac{5}{342}=0.0146 moles$
Now, molality can be calculated as:
For sucrose solution = $\dfrac{0.0146}{0.095}=0.154 m$
For glucose solution = $\dfrac{0.028}{0.095}=0.29 m$

Now, ${{K}_{f}}$ can be calculated as,
${{K}_{f}}=\dfrac{\Delta {{T}_{f}}}{molality}=\dfrac{2.15}{0.154}=13.97$
Thus,
For glucose solution, $\Delta {{T}_{f}} = {{K}_{f}}\times m=13.97\times 0.29 = 4.08$
Thus, freezing point of 5% of glucose solution in water is given as,
273.15 – 4.08 = 269.07 K

Note: For the given illustration, a theoretical equation was stated while describing the depression in freezing point where we include Van't Hoff’s factor. Do note that the solutes present here are sucrose and glucose whose molecules never dissociate. So, we can clearly ignore Van't Hoff’s factor.
Here, molality is included while solving the question. The standard units included must be checked properly and calculated accordingly.