Question

Ammonium sulfate (NH4)2SO4) is an inorganic salt with a number of commercial uses. The most common use is as a soil fertilizer. The density of a (12.5) molal (NH4)2SO4) solution is (1.77) g/mL at (25 °C). Calculate:

The mole fraction of (NH4)2SO4.
a) 0.1837
b) 0.8371
c) 1.77
d) 12.5

The molarity of (NH4)2SO4.
a) 8.35 M
b) 2.45 M
c) 1.77 M
d) 12.5 M

The percentage by mass of (NH4)2SO4.
a) 62.282%
b) 28.5%
c) 15.6%
d) 1.77%

Answer 1

To calculate the mole fraction, molarity, and percentage by mass of (NH4)2SO4 in a solution, we need to use the formulas for each calculation. The correct answers are: The mole fraction of (NH4)2SO4 is 0.1837, The molarity of (NH4)2SO4 is 8.35 M, and The percentage by mass of (NH4)2SO4 is 62.282%.

Step-by-step explanation:

To calculate the mole fraction of (NH4)2SO4, we need to know the mole fraction of the solute and the solvent. In this case, the solute is (NH4)2SO4 and the solvent is water. The mole fraction of (NH4)2SO4 can be calculated using the formula:

Mole fraction of solute = Moles of solute / Total moles of solution

The molarity of (NH4)2SO4 can be calculated using the formula:

Molarity = Moles of solute / Volume of solution (in liters)

The percentage by mass of (NH4)2SO4 can be calculated using the formula:

Percentage by mass = (Mass of solute / Mass of solution) * 100

Using the given information, we can calculate:

Mole fraction of (NH4)2SO4 = 12.5 mol / (12.5 mol + 1000 g * (1 mL / 1.77 g))

Molarity of (NH4)2SO4 = 12.5 mol / (1000 mL * (1 L / 1000 mL))

Percentage by mass of (NH4)2SO4 = (12.5 mol * 132.14 g / mol) / (12.5 mol * 132.14 g / mol + 1000 g * (1 mL / 1.77 g)) * 100

The correct answers are:

The mole fraction of (NH4)2SO4 is 0.1837.

The molarity of (NH4)2SO4 is 8.35 M.

The percentage by mass of (NH4)2SO4 is 62.282%.