Answer
a. -1.58°C.
b. -2.69°C.
c. -8.9°C.
d. -4.4°C.
Work Step by Step
We know $\Delta$t = (molality x R )
R= 1.86 C kg/mol
a. Given molality =0.85 M
$\Delta$t= (0.85 M x 1.86 C kg/mol)= 1.58°C
So, freezing point of water = 0-1.58 C= -1.58°Celsius.
b. Given molality =1.45 M
$\Delta$t= (1.45 M x 1.86 C kg/mol)= 2.69°C
So, freezing point of water = 0-2.69 C= -2.69°Celsius.
c. Given molality =4.8 M
$\Delta$t= (4.8 M x 1.86 C kg/mol)= 8.9°C
So, freezing point of water = 0-8.9 C= -8.9°Celsius.
d. Given molality =2.35 M
$\Delta$t= (2.35 M x 1.86 C kg/mol)= 4.4°C
So, freezing point of water = 0-4.4 C= -4.4°Celsius.
