The concentration of lead nitrate (Pb(NO3)2) in a 0.826 M solution is ________ molal. The density of the solution is 1.202 g/mL.
Let volume be 1 L
volume , V = 1 L
we have below equation to be used:
number of mol ,
n = Molarity * Volume
= 0.826*1
= 0.826 mol
volume , V = 1 L
= 1*10^3 mL
density, d = 1.202 g/mL
we have below equation to be used:
mass = density * volume
= 1.202 g/mL *1*10^3 mL
= 1202.0 g
This is mass of solution
Molar mass of Pb(NO3)2 = 1*MM(Pb) + 2*MM(N) + 6*MM(O)
= 1*207.2 + 2*14.01 + 6*16.0
= 331.22 g/mol
we have below equation to be used:
mass of Pb(NO3)2,
m = number of mol * molar mass
= 0.826 mol * 331.22 g/mol
= 2.736*10^2 g
This is mass of solute
mass of solvent = mass of solution - mass of solute
= 1202 - 273.5877
= 928.4123 g
= 0.92841 Kg
mass of solvent = 0.92841 Kg
we have below equation to be used:
Molality,
m = number of mol / mass of solvent in Kg
=(0.826 mol)/(0.92841 Kg)
= 0.890 molal
Answer: 0.890 molal
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