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

Given,
Molarity of the solution = 1 M
Density of the solution   = 1.25 g/mL
Mass of NaNO3 in 1L of the solution = 1 x 65 = 65 g
Mass of 1 L of the solution = 1000 x 1.25
= 1270 g
Thus,
Mass of water in the solution = 1250 - 65
= 1185 g
Hence,
Molality = Number of moles of solute / Mass of solvent in kg
= 1 / 1.185 = 0.844 m

• -5

refer to the last solved example of ncert chapter 1

• -8

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

• -10

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

• 64

thanx guyz!!!!!!!!

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