Question

Calculate the molality of 1 M solution of sodium nitrate .The density of the solution is 1.25 g/cm.

Answer 1

1.00 L of the 1.00 M solution contains 1.00 mole of NaNO3 and this is 84.994 g. 

The total mass of the solution is this: 

1000 cm^3 times 1.25 g/cm^3 = 1250 g 

How much water is in the solution: 

1250 g minus 85.0 g = 1165 g 

molality: 

1.00 mol / 1.165 kg = 0.86 m 

I rounded it to two sig fig, which seemed appropriate. You may include more if you wish. I also rounded off the 84.994 to 85 for convenience in subtraction.