18 g of glucose is dissolved in 1 kg of water. At what temperature will the solution boil ? ( `K_b` for water is 0.52 K kg `"mol"^(-1)`)

Explanation

`DeltaT_b = (k_bxx 1000 xxW_B)/(W_A xx M_B)`
`W_B = 18g , ""_1W_A = 1000 g , M_B = 180 g//mol` `k_b = 0.52 km^(-1)`
`DeltaT_b = (0.52 xx 1000xx18)/(1000 xx 180) = 9360/180000 = 0.052 K`
Boiling point of solution `rArr 373.15 + 0.052 = 373.202 K`