Calculate the molal concentration of lead nitrate (pb(no3)2) in a 0.726 m solution. the density of the solution is 1.202 g/ml.

Molarity is moles of solute (compound) in Litres of solution 
M = mol / litres 
Molality is moles of solute in kg of solvent (not solution, solvent only) 
m = mol / kg 

To work this out I first pretended I had 100 ml. Work out how many moles of Pb(NO3)2 are in 100 ml of 0.726 M Pb(NO3)2 

M = mole / L 
therefore mole = M x L 
moles Pb(NO3)2 = 0.726 M x 0.1 L 
= 0.0726 moles in 100 ml 

You are told the density is 1.202 g/ml, therefore 
weight of 100 ml of solution = 100 ml x 1.202 g/ml 
=120.2 g 

Some of this weight is attributted by the Pb(NO3)2, there are 0.0726 moles of Pb(NO3)2 in 100 ml, so work out the mass of this using 

moles = mass / molecular weight 

molecular weight Pb(NO3)2 = 207.2 + (2 x 14.01) + (6 x 16.00) = 331.22 g/mol 
So mass of Pb(NO3)2 in 100 ml = 0.0726 mol x 331.22 g/mol 
= 24.0466 g of Pb(NO3)2 in 100 ml of 0.726M solution 

To find the mass of the water subtract mass of Pb(NO3)2 from total mass. 
120.0 g - 24.0466 g = 96.1534 g 

So in 100 ml of 0.726 M Pb(NO3)2 there are 0.0726 moles (24.0466 g) of Pb(NO3)2 and 96.1534 g of water. 

Molality = moles / kg (remember moles Solute / kg SOLVENT (water)) 
= 0.0726 moles / 0.0961534 kg 
= 0.755 m

Hence, the molal concentration or molality of the solution is 0.755 m.
Thank You.