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Nitella [24]
2 years ago
11

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)
Chemistry
1 answer:
Anastaziya [24]2 years ago
4 0

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

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