If the boiling point of water is 100°C. How much gram of NaCl is added in 500g of water to increase its boiling point by approx. 1°C?

Correct option is B. 28.12g
As we know that,
ΔTb = iKb m eqn. (1)
Given:-
ΔTb = 1
i = 2
Kb = 0.52 K × kg/mol
m = w/(58.5 × 1000/500) = w/29.25
Therefore,From equation(1), we have:
1 = 2 × 0.52 × w/29.25
→ w = 28.12 gm
Hence, option B is correct.