A2.850×10−2 m solution of nacl in water is at 20.0∘
c. the sample was created by dissolving a sample of nacl in water and then bringing the volume up to 1.000 l. it was determined that the volume of water needed to do this was 999.2 ml . the density of water at 20.0∘c is 0.9982 g/ml.

To determine the boiling point of the resulting solution, use this formula:

T soln = k*m + T water

where

k =  constant of water
m = concentration of solution

Given the concentration of the solution: 2.850 x 10^-2 m 
k = 0.512
Tb water = 100 C

T soln = 0.512* 2.85x10^-2 + 100C
          = 100.0146 C

The answer for this question is d. none of these