Question

) A 0.907 M lead II nitrate solution has a density of 1.252 g/cm3. Find the molality of the solution.

Answer 1

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

$Molarity=\frac{n\times 1000}{V_s}$

where,

n= moles of solute

$V_s$ = 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 = $Density\times volume=1.252\times 1000 ml=1252g$

mass of solute =$moles\times {\text {Molar mass}}=0.907mol\times 331g/mol=300g$

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.

$Molality=\frac{n\times 1000}{W_s}$

where,

n = moles of solute  = 0.907 moles

$W_s$= weight of solvent in g  = 952 g

$Molality = \frac{0.907\times 1000}{952g}=0.953m$

Thus molality of the solution is 0.953 m