Question

A solution of $1.25g$ a non-electrolyte in $20g$ of water freezes at $271.94K$. If ${K_1} = 1.86Kmo{l^{ - 1}}$ then the molecular weight of the solute is.
1) $207.8g/mol$
2) $176.79g/mol$.
3) $209.6g/mol$.
4) $96.01g/mol$.

Answer 1

Hint:We know that Freezing point misery is the wonder that depicts why adding a solute to a dissolvable prompts the bringing down of the liquefying purpose of the dissolvable. At the point when a substance begins to freeze, the atoms hamper because of the abatements in temperature, and hence the intermolecular powers begin to need over. The freezing point depression is given by,
$\Delta {T_f} = {T_f} - T_f^* = - {K_f}m$
The freezing point depression constant is denoted as ${K_f}$.
m is the molality.

Complete answer:
Given,
The mass of the solution is $1.25g$.
The freezing point depression constant is ${K_1} = 1.86Kmo{l^{ - 1}}$.
The mass of the solvent is $20g$.
Substituting the values in the above equation,
$271.94K - 273.15K = - \dfrac{{1.86K}}{{mol}} \times m$
$m = 0.65molal$
The molecular weight of the solution can be calculated as,
By using the mole calculating formula the moles of the solute is determined.
Moles of the solute$ = \dfrac{{0.65mol}}{{Kgsolvent}}.\dfrac{{1kg}}{{1000g}} \times 20g$
Moles of the solute$ = 0.013mole$
By dividing the mass of the solution with the number of moles of solution we can determine the molecular weight of the substance.
The molecular weight$ = \dfrac{{1.25g}}{{0.013}} = 96.15g/mol$
The obtained answer is closely related to option 4.
Thus option 4 is correct.

Note:
We have to know that the particles will at that freezing point organize them during an example, and accordingly become strong. For example, as water is cooled to the softening point, its atoms become increasingly slow, bonds start to stick more, inevitably making a strong bond. On the off chance that salt is added to the water and particles pull in to the water atoms and meddle with the arrangement of the enormous organization strongly alluded to as ice. To understand a strong, the appropriate response must be cooled to a decent lower temperature.