Question

19. The solubility of ammonium nitrate in water at 25°C is 118 g per 100 mL of water and the density of a saturated aqueous solution of ammonium nitrate is 1.296 g/mL. Determine the concentration of a saturated aqueous ammonium nitrate solution in the following units.

a) Molarity

b) Molality

c) Mole Fraction

d) Weight Percent

e) ppm

25. A 0.944 M solution of glucose in water has a density of 1.0624 g/mL at 20°C. What is the concentration of glucose in this solution in the following units?

a) Mole fraction

b) Weight percent

c) Molality

Answer 1

19)

Solution -

Density of ammonium nitrate solution = 1.296 g/mL

Amount of Ammonium nitrate =118 g

Amount of water= 100 mL

a) Molarity

Number of moles per liter of solution is called as Molarity ,

Here molar mass of ammonium nitrate = 80.043 g/mol

We have to calculate moles of ammonium nitrate by using molar mass and mass ,

Number of moles (n) = (mass /molar mass )

= (118 g / 80.043 g/mol)

=1.4742 mol of ammonium nitrate ,

# amount of solution =100mL , but we need solution amount in liter so dividing by 1000,

(100/1000)=0.1 L

Molarity = (moles /per liter of solution )

=(1.4742 mol /0.1 Lit)

=14.74M ammonium nitrate solution

b) Molality

Molality (m)= (moles of solute /kg of solvent)

First we have to determine mass of solvent in kg by using density of solvent-

Amount of solvent (water ) = 100mL , we know that density of water is 1.0 g/mL ,so mass of solvent = 100 gm

But we need it in Kg,

100 g x (1kg/1000 g) = 0.1 kg of water

We already calculated moles of ammonium nitrate = 1.4742 mol,

Molality (m) =( mol/ kg of solvent )

= (1.4742 mol / 0.1 kg)

= 14.74 mol /kg

Molality (m) = 14.74 m

C)mole fraction

Mole fraction is defined as amount of constituents divide by total number of constituents in solution .

Mole fraction of solute =

X_solute = (no. of moles of solute) / (no. of moles of solute + no. of moles of solvent)

Here solute is ammonium nitrate and it's calculated moles are 1.4742 mol , while moles of solvent (water )need to determine -

#molar mass of water =18.0g/mol

Moles of water = (mass of water /molar mass of water )

= (100 g /18 g/mol)

=5.555 mol of water (solvent )

#Mol fraction of solute =

X _solute= ( 1.4742 )/(1.4742 +5.555)

=0.20

Mole fraction of solute (X _solute)= 0.20

One fact is clear that

X _solute + X _solvent = 1

By putting values ,

0.20 + X _solvent = 1

So, X _solvent = (1.0 - 0.20)

=0.80

D) weight percent

#Here mass of ammonium nitrate ( solute) =118 gm

# mass of solution = mass of solute + mass of solvent

= 118 g + 100 g

=218 g of solution

% by mass of solute ,

% w/w = (mass of solute/mass of solution ) x 100

= (118 /218 ) x100

=54.12 %

E) ppm

Part per million is gram of solute per million gram of solution

Here total volume of solution =100ml

But density of solution = 1.296g/mL

So mass of solution = Volume  x density

= 100 mL x 1.296 g/mL

=129.6 g of solution

ppm =(mass of solute /mass of solution ) x 10⁶

=(118 / 129.6) x 10⁶

=0.91 x 10⁶ ppm