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1% lead nitrate = 1g lead nitrate in 100g water. ( This is technically incorrect I know, but it is the simplest way of doing this problem)
Solution = 10g Pb(NO3)2 in 1.0kg water
Molar mass Pb(NO3)2 = 331.2200 g/mol
10g Pb(NO3)2 = 10/331.22 = 0.030 mol in 1kg solvent
solution is 0.03 molal
Kf for water = 1.86°C /m.kg
van't Hoff factor for Pb(NO3)2, i = 3
molality = 0.03m
delta T = i*Kf*m
delta t = 3*1.86*0.03
delta T = 0.168°C
The solution will freeze at - 0.168°C.
Solution = 10g Pb(NO3)2 in 1.0kg water
Molar mass Pb(NO3)2 = 331.2200 g/mol
10g Pb(NO3)2 = 10/331.22 = 0.030 mol in 1kg solvent
solution is 0.03 molal
Kf for water = 1.86°C /m.kg
van't Hoff factor for Pb(NO3)2, i = 3
molality = 0.03m
delta T = i*Kf*m
delta t = 3*1.86*0.03
delta T = 0.168°C
The solution will freeze at - 0.168°C.
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