Question

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.

Answer 1

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