Question

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

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 :  

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

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

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

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

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