Calculate the molality of 1M solution of sodium nitrate. The Density of the solution is 1.25g/ml.

refer to the last solved example of ncert chapter 1

given,molality of 1M and the density of sodium nitrate is 1.25 g/ml

we know that molarity=moles of solute/volume of solution in litre

molarity=1M , then density=mass/volume

mass of sodium nitrate =23*2+14+16*3=108

  1.25 =108/volume of solution in liter

volumeof solutionin litre =108/1.25

  = 86.4litre

                moles of solute=86.4

  m(molarity)           =86.4/86.5

                                            =1m 

M = no. of moles of solute / vol .of solution (L)

m = no. of moles of solute / Wt. of solvent (Kg)

1M sodium nitrate solution contains 85 g of sodium nitrate salt (23+14+48)

From the density it is well known that 1 L of solution  weight is 1250 g

so wt. of water present = 1250-85 = 1165 g = 1.165 Kg

Hence the molality = 1/1.165 = 0.858

thanx guyz!!!!!!!!