Find booling point of 2.5% W/W sugar solution,the molal elevation constant for solvent is 4K kg/mole.

Dear Student 

According to colligative property: 

The elevation in boiling point directly depends upon the number of solutes. 

$$\begin{gathered} ∆T_b=K_b\times m \\ [Here ∆T_b = elevation in boiling point ; \\ {K}_b= ebulioscopic cons\tan t or molal elevation point ; \\ m = molality] \end{gathered}$$

So the molality of the solution: 
          $$\begin{gathered} =\frac{No. of moles of solute }{Weight of solvent in \left(g\right) }\times 1000 \\ = \frac{2.5}{342 \times 100}\times 1000 \\ =0.073 \end{gathered}$$

Now the elevation of boiling point : 
         $$\begin{gathered} =4 \times 0.073 \\ =0.292 °K \end{gathered}$$

So, as for the normal water the boiling point is 373^o K. 
The elevated boiling point of the solution is = 373 + 0.292 
                                                                       = 373.292^oK

Regards