Find The boiling point of water at 735 torr is 99.07 the mass of nacl added in 100 g of water to makes its boiling point 100 celcius (Kb=0.51)
Dear student,
Elevation in boiling point , ΔTb = (100 + 273) - ( 99.07 + 273)
= 0.93 K
Mass of water, w1 = 100 g
Molar mass of NaCl, M2 = 58.5 g mol-1
Molal elevation constant, Kb = 0.51 kg K mol-1
ΔTb =
Thus, mass of NaCl to be added = 10.667 g
Regards
Elevation in boiling point , ΔTb = (100 + 273) - ( 99.07 + 273)
= 0.93 K
Mass of water, w1 = 100 g
Molar mass of NaCl, M2 = 58.5 g mol-1
Molal elevation constant, Kb = 0.51 kg K mol-1
ΔTb =
Thus, mass of NaCl to be added = 10.667 g
Regards