Chemistry, 02.09.2019 18:20 mbarua

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

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Answer from: walterzea70

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.

Answer from: Quest

"the periodic table is arranged in order of atomic number."

Answer from: Quest

the population of unionid mussels in the hudson river decreased after zebra mussels were introduced. zebra mussels, a nonnative species, competed with the unionid mussels for food. the zebra mussels were able to outcompete the unionid mussels, allowing the zebra mussel population to increase dramatically over six years. the unionid mussel population decreased due to a lack of food.

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