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 :

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

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 = ?

$T^o_b$ = 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:

$(T_b-100)^oC=1\times (0.52^oC/m)\times \frac{(5.0g)\times 1000}{342.3\times (10.0g)}$

$T_b=100.6^oC$

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