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


ΔTbKb×1000×w2M2×w1

w2 =Tb×M2×w1Kb×1000            = 0.93×58.5×1000.51×1000              = 10.667 g

Thus, mass of NaCl to be added = 10.667 g

Regards

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