Question

A $5\% $ solution (by mass) of urea in water has a freezing point of $271.52K$ . Calculate the freezing point of $5\% $ glucose in water if the freezing point of pure water is $273.15K$.

Answer 1

Hint: To solve this problem, firstly we should understand concepts like Molar mass, Molality and Freezing point and study the concept behind these terms. We will also study the equations and formulas required to solve this question, and then by following these steps, we can easily approach our answer. So, let's study them in brief-



Complete step by step solution: Molar mass- Molar mass is defined as the smallest mass unit of a compound with one twelfth of the mass of one carbon-12 atom. The standard unit for this is $g\,mo{l^{ - 1}}$
Molality- Molality states that the total moles of a solute contained in a kilogram of a solvent. Molality is also known as molal concentration. It is an estimate of solute concentration in a solution. The solution consists of two components; solute and solvent.
                                            Molality$\left( M \right) = \dfrac{{Number\,of\,moles\,of\,solute}}{{Mass\,of\,solvent\,in\,kgs}}$
                                           $Molality\left( M \right) = \dfrac{{g \times 1000}}{{W \times m}}$
Freezing point is the temperature of a liquid at which it alters its state from liquid to solid at atmospheric pressure. At freezing point, these two phases liquid and solid prevail in equilibrium $i.e\,\,$ at this point both solid state and liquid state exist simultaneously. The freezing point of a substance looks upon atmospheric pressure.
Freezing point of a substance is the temperature of that substance at which the substance starts to freeze.
For example- Freezing point of Water $ = {0^ \circ }C$
                                                             $ = 273.15K$

Given,
Molar mass of Urea$ = 60g/mol$
 Molality$\left( m \right)$ of Urea $ = \dfrac{{5 \times 1000}}{{60 \times 100}}$
$ \Rightarrow 0.833m$
$\Delta {T_f}$ for solution$ = 2.15K$
$\Delta {T_f} = {K_f} \times m$
$ \Rightarrow {K_f} = \dfrac{{\Delta {T_f}}}{m} = \dfrac{{2.15}}{{0.83}} = 2.6$

Now, Molality of Glucose Solution
$ = \dfrac{5}{{180}} \times \dfrac{{1000}}{{100}} = 0.278$
$\Delta {T_f} = 2.6 \times 0.278$
$ = 0.7228K$

Freezing point of Glucose solution
$ = 273.15 - 0.722$
$ = 272.42K$



Note: Freezing refers to the phase change of a substance from the liquid state to a solid- state. This is a phase change phenomenon which is transformed from one state of matter to another state. The freezing point of solid occurs at the temperature in which the solid and liquid phases are in equilibrium.