Question

Benzyl acetate is one of the active components of oil of jasmine. If 0.125 g of the compound is added to 25.0g of chloroform (CHCl 3 ), the boiling point of the solution is 61.82°C. What is the molar mass of benzyl acetate. (Chloroform: Kb = 3.63 °C/m; Tb° 61.70°C)

Answer 1

Answer : The molar mass of benzyl acetate is 151.25 g/mol

Explanation :

Formula used for Elevation in boiling point :

`\Delta T_b=i* k_b* m`

or,

`T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* w_1)`

where,

`T_b` = boiling point of solution =
`61.82^oC`

`T^o_b` = boiling point of chloroform =
`61.70^oC`

`k_b` = boiling point constant of chloroform =
`3.63^oC/m`

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

`w_2` = mass of solute (Benzyl acetate) = 0.125 g

`w_1` = mass of solvent (chloroform) = 25.0 g

`M_2` = molar mass of solute (Benzyl acetate) = ?

Now put all the given values in the above formula, we get:

`(61.82-61.70)^oC=1* (3.63^oC/m)* ((0.125g)* 1000)/(M_2* (25.0g))`

`M_2=151.25g/mol`

Therefore, the molar mass of benzyl acetate is 151.25 g/mol