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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)

Respuesta :

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

Explanation:

Depression in freezing point :

Formula used for lowering in freezing point is,

[tex]\Delta T_f=k_f\times m[/tex]

or,

[tex]\Delta T_f=k_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}[/tex]

where,

[tex]T_f[/tex] = change in freezing point

[tex]k_f[/tex] = freezing point constant= [tex]9.10^0C/m[/tex]

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

[tex]\Delta T_f=T^{o}_f-T_f=25.5^0C-15.3^0C)=10.2^0C[/tex]

[tex]10.2=9.10\times \frac{0.807}{\text{molar mass of solute}\times 0.0116kg}[/tex]

[tex]{\text{molar mass of solute}}=62.07 g/mol[/tex]

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

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