find boiling point of solution containing 0.52g of glucose dissolved in 80.2 g of water. kb=0.52 Kkg/m.is the answer=373.132???

Given,
W_solvent = 80.2 g
W_solute = 0.52 g
K_b = 0.52 Kkg/m
Molar mass, M of glucose = 12 x 6 +12 x 1 + 6 x 16 =180g/mol

We know,
$$\begin{gathered} ∆T_b=\frac{K_b x W_{solute} x 1000}{W_{solvent x M_{solute}}} \\ =\frac{ 0.52 x 0.52 x 1000}{80.2 x 180} \\ = 0.0187 K \end{gathered}$$

So, the boiling point of solution = Normal boiling point of water + elevation in boiling point
  = 373.15 + 0.0187
  = 373.168 K