Question

a solution containing 2.2 grams of a substance in 125 grams of benzene boils at 80. 66 degree Celsius kb for 1 kg of benzene is 33. 28 degree Celsius/ molal, the molecular weight of the substance shall be

Answer 1

Answer : The molecular weight of the substance will be, 130.7 g/mol

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m$

or,

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = $80.66^oC$

$T^o_b$ = boiling point of benzene = $80.1^oC$

$k_b$ = boiling point constant  of benzene for 1 kg = $33.28^oC/m$

Boiling point constant  of benzene for 125 g = $\frac{125}{1000}\times 33.28=4.16^oC/m$

m = molality

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

$w_2$ = mass of solute = 2.2 g

$w_1$ = mass of solvent (benzene) = 125 g

$M_2$ = molar mass of solute = ?

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

$(80.66-80.1)^oC=1\times (4.16^oC/m)\times \frac{(2.2g)\times 1000}{M_2\times (125g)}$

$M_2=130.7g/mol$

Therefore, the molecular weight of the substance will be, 130.7 g/mol