Answer:
Explanation:
From the given information:
TO start with the molarity of the solution:
[tex]= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}[/tex]
= 0.601 mol/kg
= 0.601 m
At the freezing point, the depression of the solution is [tex]\Delta \ T_f = T_{solvent}- T_{solution}[/tex]
[tex]\Delta \ T_f = 2.9 ^0 \ C[/tex]
Using the depression in freezing point, the molar depression constant of the solvent [tex]K_f = \dfrac{\Delta T_f}{m}[/tex]
[tex]K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}[/tex]
[tex]K_f = 4.82 ^0 C / m}[/tex]
The freezing point of the solution [tex]\Delta T_f = T_{solvent} - T_{solution}[/tex]
[tex]\Delta T_f = 7.3^ 0 \ C[/tex]
The molality of the solution is:
[tex]= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}[/tex]
Molar depression constant of solvent X, [tex]K_f = 4.82 ^0 \ C/m[/tex]
Hence, using the elevation in boiling point;
the Vant'Hoff factor [tex]i = \dfrac{\Delta T_f}{k_f \times m}[/tex]
[tex]i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}[/tex]
[tex]\mathbf {i = 1.51 }[/tex]
