Answer
a. -0.18°C.
b. -0.87°C.
c. -2.6°C.
d. -11°C.
Work Step by Step
We know $\Delta$t= (molality x R )
R= 1.86 C kg/mol
a. Given molality =0.100M
$\Delta$t = (0.100m x 1.86 C kg/mol)= 0.18 C
So, freezing point of water = 0-0.18 C= -0.18°C.
b. Given molality =0.469 M
$\Delta$t= (0.469m x 1.86 C kg/mol)= 0.87 C
So, freezing point of water = 0-0.87 C= -0.87°C.
c. Given molality =1.44 M
$\Delta$t= (1.44m x 1.86 C kg/mol)= 2.6 C
So, freezing point of water = 0-2.6C= -2.6°C.
d. Given molality =5.89M
$\Delta$t= (5.89m x 1.86 C kg/mol)= 11 C
So, freezing point of water = 0-11 C= -11°C.
