Question

If 4.27 g sucrose (c12h22o11) are dissolved in 15.2 g water, what is the boiling point of the resulting solution? kb for water = 0.512c/m.

Answer 1

Answer is: the boiling point of the resulting solution of sucrose is 100.42°C.
m(H₂O) = 15.2 g ÷ 1000 g/kg = 0.0152 kg.
m(C₁₂H₂₂O₁₁) = 4.27 g. 
n(C₁₂H₂₂O₁₁) = m(C₁₂H₂₂O₁₁) ÷ M(C₁₂H₂₂O₁₁).
n(C₁₂H₂₂O₁₁) = 4.27 g ÷ 342.3 g/mol.
n(C₁₂H₂₂O₁₁) = 0.0125 mol.
b(solution) = n(C₁₂H₂₂O₁₁) ÷ m(H₂O).
b(solution) = 0.0125 mol ÷ 0.0152 kg.
b(solution) = 0.82 m.
ΔT = b(solution) · Kb(H₂O).
ΔT = 0.82 m · 0.512°C/m.
ΔT = 0.42°C.
Tb = 100°C + 0.42°C = 100.42°C.

Answer 2

Answer : The boiling point of a solution is, $100.42^oC$

Explanation :

Formula used for Elevation in boiling point :

$\Delta T_b=k_b\times m$

or,

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

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure water = $100^oC$

$k_b$ = boiling point constant  for water = $0.512^oC/m$

m = molality

$w_2$ = mass of solute (sucrose) = 4.27 g

$w_1$ = mass of solvent (water) = 15.2 g

$M_2$ = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in the above formula, we get the boiling point of a solution.

$T_b-100^oC=\frac{1000\times 0.512^oC/m\times 4.27g}{15.2g\times 342.3g/mole}$

$T_b=100.42^oC$

Therefore, the boiling point of a solution is, $100.42^oC$