Question

When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2.9 °C lower than the freezing point of pure X. On the other hand, when 61.6 g of ammonium chloride (NH4CI) are dissolved in the same mass of X. The freezing point of the solution is 7.3 °C lower than the freezing point of pure X Calculate the van't Hoff factor for ammonium chloride in X.

Answer 1

Step-by-step explanation:

From the given information:

TO start with the molarity of the solution:

`= (61.6 \ g * (1 \ mol \ C_3H_7 NO_3)/(89.1 \ g) )/(1150 \ g * (1 \ kg)/(1000 \g ))`

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is
`\Delta \ T_f = T_(solvent)- T_(solution)`

`\Delta \ T_f = 2.9 ^0 \ C`

Using the depression in freezing point, the molar depression constant of the solvent
`K_f = (\Delta T_f)/(m)`

`K_f = (2.9 ^0 \ C)/(0.601 \ m)`

`K_f = 4.82 ^0 C / m}`

The freezing point of the solution
`\Delta T_f = T_(solvent) - T_(solution)`

`\Delta T_f = 7.3^ 0 \ C`

The molality of the solution is:

`= (61.6 \ g * (1 \ mol \ NH_4Cl)/(53.5 \ g) )/(1150 \ g * (1 \ kg)/(1000 \g ))`

Molar depression constant of solvent X,
`K_f = 4.82 ^0 \ C/m`

Hence, using the elevation in boiling point;

the Vant'Hoff factor
`i = (\Delta T_f)/(k_f * m)`

`i = (7.3 \ ^0 \ C)/(4.82 ^0 \ C/m * 1.00 \ m)`

`\mathbf {i = 1.51 }`