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19. The solubility of ammonium nitrate in water at 25°C is 118 g per 100 mL of w

ID: 710837 • Letter: 1

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

Explanation / Answer

19. Accoridng to given data:

The volume of water (V) = 100 mL = (100/1000) L = 0.1 L

The density of saturated aqueous NH4NO3 = 1.296 g/mL

Therefore, the mass of NH4NO3 in saturated aqueous solution = 100 mL * 1.296 g/mL = 129.6 g

a) Molarity of saturated aqueous NH4NO3 = nNH4NO3/V ( in L), where n = corresponding no. of moles

Here, the molar mass of NH4NO3 = 80 g/mol

i.e. nNH4NO3 = 129.6 g/80 g mol-1 = 1.62 mol

Therefore, molarity = 1.62 mol/0.1 L = 16.2 M

b) molality of saturated aqueous NH4NO3 = nNH4NO3/Mass of solution ( in Kg)

The density of water at 25 oC = 1 g/mL

i.e. 100 mL water = 100 g water = (100/1000) Kg = 0.1 Kg

Therefore, molality = 1.62 mol/0.1 Kg = 16.2 m

c) molefraction of NH4NO3 = nNH4NO3/n, where n = total no. of moles = nNH4NO3 + nH2O

Here, nH2O = 100 g/18 g mol-1 = 5.556 mol

Now, n = 1.6 2 + 5.556 = 7.176

Therefore, molefraction = 1.62 mol/7.176 mol = 0.226

d) Weight percent of NH4NO3 in the saturated aqueous solution = (mass of NH4NO3/total mass of solution)*100

Here, the total mass of solution = 129.6 g + 100 g = 229.6 g

Now, the weight percent of NH4NO3 = (129.6 g/229.6)*100 = 56.446%

e) Now, the concentration of NH4NO3 in the saturated aqueous solution = 129.6 g/0.1 L

= 129.6*103 mg/0.1 L

= 1.296*106 mg/L (Note: 1 ppm = 1 mg/L)

= 1.296*106 ppm

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