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

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

Step-by-step explanation:

`\Delta T_f=K_f* 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`=
`(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=(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](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'=(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`=
`(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''=(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