Question

The freezing point of ethanol, CH3CH2OH, is -117.300 °C at 1 atmosphere. Kf(ethanol) = 1.99 °C/m

In a laboratory experiment, students synthesized a new compound and found that when 12.70 grams of the compound were dissolved in 216.5 grams of ethanol, the solution began to freeze at -117.431 °C. The compound was also found to be nonvolatile and a non-electrolyte.

What is the molecular weight they determined for this compound?

Answer 1

Answer : The molecular weight of this compound is 891.10 g/mol

Explanation : Given,

Mass of compound = 12.70 g

Mass of ethanol = 216.5 g

Formula used :

`\Delta T_f=i* K_f* m\\\\T_f^o-T_f=i* T_f*\frac{\text{Mass of compound}* 1000}{\text{Molar mass of compound}* \text{Mass of ethanol}}`

where,

`\Delta T_f` = change in freezing point

`T_f^o` = temperature of pure ethanol =
`-117.300^oC`

`T_f` = temperature of solution =
`-117.431^oC`

`K_f` = freezing point constant of ethanol =
`1.99^oC/m`

i = van't hoff factor = 1 (for non-electrolyte)

m = molality

Now put all the given values in this formula, we get

`(-117.300)-(-117.431)=1* 1.99^oC/m* \frac{12.70g* 1000}{\text{Molar mass of compound}* 216.5g}`

`\text{Molar mass of compound}=891.10g/mol`

Therefore, the molecular weight of this compound is 891.10 g/mol