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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