Question

What is the boiling point (in °C) of a solution of 7.94 g of I2 in 69.2 g of toluene, assuming the I2 is nonvolatile? (For toluene, Tb = 110.63°C and Kb = 3.40°C·kg/mol.)

Answer 1

Answer: The boiling point of  solution is $112.16^0C$

Explanation:

Elevation in boiling point:

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

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of toluene = $110.63^oC$

$k_b$ = boiling point constant of toluene =$3.40^oC/m$

m = molality

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

$w_2$ = mass of solute $(I_2)$ = 7.94 g

$w_1$ = mass of solvent (toluene) = 69.2 g

$M_2$ = molar mass of solute $(I_2)$= 254g/mol

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

$(T_b-110.63)^oC=1\times (3.40^oC/m)\times \frac{(7.94g)\times 1000}{254\times (69.2g)}$

$T_b=112.16^0C$

Therefore, the boiling point (in °C) of a solution is 112.16