Question

When 36.0 g of a solute having the empirical formula $C{H_2}O$ is dissolved in 1.20 Kg of water, the solution freezes at \[ - {0.93^\circ }C\]. what is the molecular formula of the compound?
A. ${C_3}{H_5}{O_2}$
B. ${C_3}{H_4}{O_3}$
C. ${C_2}{H_4}{O_2}$
D. ${C_3}{H_4}{O_2}$

Answer 1

Hint: We know the freezing point depression, $\Delta {T_f} = {k_f} \times m$ , and ${k_f}$ for water \[ = \dfrac{{{{1.86}^\circ }C}}{m}\] . We also know Molality = moles of solute / kg of solvent. Thus, we can calculate the molar mass of solute. And, since we have the empirical formula, we can calculate the value of ‘n’.

Complete step by step answer:
Given,
Mass of solute = 36.0 g
Mass of solvent = 1.20 Kg
We know, $\Delta {T_f} = {k_f} \times m$
Where, $\Delta {T_f} = $ Change in Freezing point
 ${k_f} = $ Molal freezing point depression constant of the solvent
m = molality
 $\Delta {T_f} = {k_f} \times m$
 \[ \Rightarrow 0.93 = 1.83 \times m\] …. equation 1
( ${k_f}$ for water \[ = \dfrac{{{{1.86}^\circ }C}}{m}\] )
Molality = m = moles of solute / kg of solvent
Molality $ = \dfrac{{Mass\,of\,solute}}{{Molar\,mass\,of\,solute \times mass\,of\,solvent}}$
(Since, Number of moles $ = \dfrac{{mass}}{{molar\,mass}}$ )
So, we get
 $m = \dfrac{{36}}{{x \times 1.2}}$ …. equation 2
Putting the value of m from equation 2 in equation 1
 $ \Rightarrow 0.93 = 1.83 \times \dfrac{{36}}{{x \times 1.2}}$
 \[ \Rightarrow x = 60\]
Molar mass of solute = 60
Given empirical formula $ = C{H_2}O$
Empirical molecular mass \[ = 12 + \left( {1 \times 2} \right) + 16 = 30\]
Now, we can find the value of ‘n’
 \[ \Rightarrow n = \dfrac{{60}}{{30}} = 2\]
Molecular formula = (n) $ \times $ (Empirical formula)
So, molecular formula becomes $ = {C_2}{H_4}{O_2}$

Therefore, the correct answer is option (C).

Note: The freezing point of the pure solvent is at a constant temperature but the freezing point of the solution slowly decreases. This decrease is caused by the increase in the solute concentration as the solvent freezes. The equation which describe the change in freezing point from pure solvent to solution is, $\Delta {T_f} = {k_f} \times m$ .