Answer: The molality of the solution is 0.953 m
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution
[tex]Molarity=\frac{n\times 1000}{V_s}[/tex]
where,
n= moles of solute
[tex]V_s[/tex] = volume of solution in ml
Given : 0.907 moles of lead(ii)nitrate is dissolved in 1000 ml of the solution.
density of solution= 1.252 g/ml
Thus mass of solution = [tex]Density\times volume=1.252\times 1000 ml=1252g[/tex]
mass of solute =[tex]moles\times {\text {Molar mass}}=0.907mol\times 331g/mol=300g[/tex]
mass of solvent =mass of solution - mass of solute = (1252-300)g= 952 g
Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.
[tex]Molality=\frac{n\times 1000}{W_s}[/tex]
where,
n = moles of solute = 0.907 moles
[tex]W_s[/tex]= weight of solvent in g = 952 g
[tex]Molality = \frac{0.907\times 1000}{952g}=0.953m[/tex]
Thus molality of the solution is 0.953 m
