Question

tert-Butyl alcohol is a solvent with a Kf of 9.10 ∘C/m and a freezing point of 25.5 ∘C. When 0.807 g of an unknown colorless liquid was dissolved in 11.6 g of tert-butyl alcohol, the solution froze at 15.3 ∘C.Which of the following is most likely the identity of this unknown liquid?tert-Butyl alcohol is a solvent with a of 9.10 and a freezing point of 25.5 . When 0.807 of an unknown colorless liquid was dissolved in 11.6 of tert-butyl alcohol, the solution froze at 15.3 .Which of the following is most likely the identity of this unknown liquid?ethylene glycol (molar mass = 62.07 g/mol)1-octanol (molar mass = 130.22 g/mol)glycerol (molar mass = 92.09 g/mol)2-pentanone (molar mass = 86.13 g/mol)1-butanol (molar mass = 74.12 g/mol)

Answer 1

Answer: ethylene glycol (molar mass = 62.07 g/mol)

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

`\Delta T_f=k_f* m`

or,

`\Delta T_f=k_f* \frac{\text{mass of solute}}{\text{molar mass of solute}* \text{weight of solvent in kg}}`

where,

`T_f` = change in freezing point

`k_f` = freezing point constant=
`9.10^0C/m`

m = molality

Given mass of solute = 0.807 g

Molar mass of solute=? g/mol

weight of solvent in kg= 11.6 g=0.0116 kg

`\Delta T_f=T^(o)_f-T_f=25.5^0C-15.3^0C)=10.2^0C`

`10.2=9.10* \frac{0.807}{\text{molar mass of solute}* 0.0116kg}`

`{\text{molar mass of solute}}=62.07 g/mol`

Thus the solute is ethylene glycol which has same molecular mass as calculated, i.e 62.07 g/mol.