The boiling point of water at 735 torr is 99.07°C. The mass of NaCl added in 100g water to make its boiling point 100°C is?(k=0.51 k kg/mol)
Step 1:
Given that, kb=0.51KKgmol−1
ΔTf=100−99.07=0.93oC
Step 2:
Using the relation, ΔTf=Kf×m
m=0.930.51=1.824
nNaClWH2O=1.824
Step 3:
nNaCl=0.5×1.824×100×10−3=0.0911
mass = 0.0911×58.5=5.33g
Hence, the total amount of NaCl to be added is calculated as 5.33 g.