Question

Which of the following solutions will have the lowest freezing point? Input the appropriate letter. A. 35.0 g of C3H8O in 250.0 g of ethanol (C2H5OH) B. 35.0 g of C4H10O in 250.0 g of ethanol (C2H5OH) C. 35.0 g of C2H6O2 in 250.0 g of ethanol (C2H5OH)

Answer 1

Answer:

Solution with 35.0 g of $C_3H_8O$ in 250.0 g of ethanol will have lowest freezing point

Explanation:

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =  

$K_f$ = freezing point constant  

m = molality

As we can see that higher the molality of the solution more will depression in freezing point of the solution and hence lower will the freezing point of solution.

$Molality=\frac{moles}{\text{mass of solvent in kg}}$

A. 35.0 g of $C_3H_8O$ in 250.0 g of ethanol.

Moles of $C_3H_8O$=$\frac{35.0 g}{60 g/mol}=0.5833 mol$

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m=\frac{0.5833 mol}{0.25 kg}=2.33 m$

B. 35.0 g of $C4H_{10}O$ in 250.0 g of ethanol

Moles of $C_4H_{10}O$$=[tex]\frac{35.0 g}{74 g/mol}=0.4730 mol$

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m'=\frac{0.4730 mol}{0.25 kg}=1.89 m$

C. 35.0 g of $C_2H_{6}O_2$ in 250.0 g of ethanol

Moles of $C_2H_{6}O_2$=$\frac{35.0 g}{62g/mol}=0.5645 mol$

Mass of solvent i.e. ethanol = 250.0 g = 0.25 kg (1 g = 0.001 kg)

$m''=\frac{0.5645 mol}{0.25 kg}=2.26 m$

$m>m'''>m''$

Solution with 35.0 g of $C_3H_8O$ in 250.0 g of ethanol will have lowest freezing point