Question

Calculate the number of moles and the mass of the solute in each of the following solutions:
(a) 325 mL of 8.23 x 10-5 M Kl, a source of iodine in the diet
(b) 75.0 mL of 22 x 10-5 M H2SO4, a sample of acid rain
(c) 0.2500 L of 0.1135 M K2CrO4 and analytical reagent used in iron assays
(d)10.5 L of 3.716 M (NH4)2SO4, a liquid fertilizer

Answer 1

Answer:

For a: The number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

For b: The number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

For c: The number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

For d: The number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$    .....(1)

To calculate the number of moles of a substance, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(2)

• For a:

Molarity of KI = $8.23\times 10^{-5}M$

Volume of solution = 325 mL = 0.325 L     (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$8.25\times 10^{-5}mol/L=\frac{\text{Moles of KI}}{0.325L}\\\\\text{Moles of KI}=2.7\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of KI = $2.7\times 10^{-5}mol$

Molar mass of KI = 166 g/mol

Putting values in equation 2, we get:

$2.7\times 10^{-5}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=4.482\times 10^{-3}g$

Hence, the number of moles of KI are $2.7\times 10^{-5}$ and mass is $4.482\times 10^{-3}g$

• For b:

Molarity of sulfuric acid = $22\times 10^{-5}M$

Volume of solution = 75 mL = 0.075 L

Putting values in equation 1, we get:

$22\times 10^{-5}mol/L=\frac{\text{Moles of sulfuric acid}}{0.075L}\\\\\text{Moles of }H_2SO_4=1.65\times 10^{-5}mol$

Now, using equation 2, we get:

Moles of sulfuric acid = $1.65\times 10^{-5}mol$

Molar mass of sulfuric acid = 98 g/mol

Putting values in equation 2, we get:

$1.65\times 10^{-5}mol=\frac{\text{Mass of }H_2SO_4}{98g/mol}\\\\\text{Mass of }H_2SO_4=1.617\times 10^{-3}g$

Hence, the number of moles of sulfuric acid are $1.65\times 10^{-5}$ and mass is $1.617\times 10^{-3}g$

• For c:

Molarity of potassium chromate = $0.1135M$

Volume of solution = 0.250 L

Putting values in equation 1, we get:

$0.1135mol/L=\frac{\text{Moles of }K_2CrO_4}{0.250L}\\\\\text{Moles of }K_2CrO_4=2.84\times 10^{-2}mol$

Now, using equation 2, we get:

Moles of potassium chromate = $2.84\times 10^{-2}mol$

Molar mass of potassium chromate = 194.2 g/mol

Putting values in equation 2, we get:

$2.84\times 10^{-2}mol=\frac{\text{Mass of }K_2CrO_4}{194.2g/mol}\\\\\text{Mass of }K_2CrO_4=5.51g$

Hence, the number of moles of potassium chromate are $2.84\times 10^{-2}$ and mass is 5.51 g.

• For d:

Molarity of ammonium sulfate = 3.716 M

Volume of solution = 10.5 L

Putting values in equation 1, we get:

$3.716mol/L=\frac{\text{Moles of }(NH_4)_2SO_4}{10.5L}\\\\\text{Moles of }(NH_4)_2SO_4=39.018mol$

Now, using equation 2, we get:

Moles of ammonium sulfate = 39.018 mol

Molar mass of ammonium sulfate = 132.14 g/mol

Putting values in equation 2, we get:

$39.018mol=\frac{\text{Mass of }(NH_4)_2SO_4}{132.14g/mol}\\\\\text{Mass of }(NH_4)_2SO_4=5155.84g$

Hence, the number of moles of ammonium sulfate are 39.018 moles and mass is 5155.84 grams.