Question

The elevation of a boiling point of a solution containing 1.0 g of a solute(mol:weight =100) in 10of water0 g

Answer 1

ΔTb = iKbmolality

molality =1/100 x1000/100 =1/10=0.1

ΔTb = 1x1.53 x0.1=0.153

Ts -T0 =0.153

Ts -100 = 0.153

Ts =100+0.153=101.53 Celsius

elevation of boling point of solution =101.53 Celsius.

Answer 2

Answer:

101 is correct answer.