Question

If 5.0 grams of sucrose, C12H22O11, are dissolved in 10.0 grams of water, what will be the boiling point of the resulting solution?

Answer 1

Answer : The boiling point of the resulting solution is,
`100.6^oC`

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 = ?

`T^o_b` = boiling point of water =
`100^oC`

`k_b` = boiling point constant =
`0.52^oC/m`

m = molality

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

`w_2` = mass of solute (sucrose) = 5.0 g

`w_1` = mass of solvent (water) = 10.0 g

`M_2` = molar mass of solute (sucrose) = 342.3 g/mol

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

`(T_b-100)^oC=1* (0.52^oC/m)* ((5.0g)* 1000)/(342.3* (10.0g))`

`T_b=100.6^oC`

Therefore, the boiling point of the resulting solution is,
`100.6^oC`