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`

Step-by-step explanation:

Elevation in boiling point:

`T_b-T^o_b=i* k_b* (w_2* 1000)/(M_2* 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* (3.40^oC/m)* ((7.94g)* 1000)/(254* (69.2g))`

`T_b=112.16^0C`

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