Answer:
-3.63 degree Celsius
Explanation:
We are given that
Boiling point of solution=[tex]T_b=101^{\circ}[/tex] C
Boiling point water=100 degree Celsius
[tex]K_f=1.86K/m[/tex]
[tex]K_b=0.512 K/m[/tex]
[tex]\Delta T_b=T-T_0[/tex]
Where [tex]T[/tex]=Boiling point of solution
[tex]T_0=[/tex]Boiling point of pure solvent
[tex]\Delta T_b=101-100=1^{\circ}[/tex]C
[tex]\Delta T_b=k_bm[/tex]
Using the formula
[tex]1=0.512\times m[/tex]
Molality,[tex]m=\frac{1}{0.512}[/tex] m
[tex]\Delta T_f=k_fm[/tex]
Using the formula
[tex]\Delta T_f=\frac{1}{0.512}\times 1.86[/tex]
[tex]\Delta T_f=3.63 C[/tex]
We know that
[tex]\Delta T_f=T_0-T_1[/tex]
Where [tex]T_0[/tex] =Freezing point of solvent
[tex]T_1=[/tex] Freezing point of solution
Using the formula
[tex]3.63=0-T_1[/tex]
Freezing point of water=0 degree Celsius
[tex]T_1=0-3.63=-3.63 C[/tex]
Hence, the freezing point of solution=-3.63 degree Celsius
