Question

A solution was prepared by dissolving 35.0 g of KCl in 225 g of water.a. Calculate the mass percent of KCl in the solution. Express your answer with the appropriate units.b. Calculate the mole fraction of the ionic species KCl in the solution. Express the concentration numerically as a mole fraction in decimal form.c. Calculate the molarity of KCl in the solution if the total volume of the solution is 239 mL. Express your answer with the appropriate units.d. Calculate the molality of KCl in the solution. Express your answer with the appropriate units.

Answer 1

Answer:

1) mass % KCl = 13.5 %

2) mol fraction KCl= 0.0362 = 3.6 * 10^-2

3) Molarity KCl =  1.96 M

4) Molality KCl = 2.08 molal

Explanation:

Step 1: Data given

Mass of KCl = 35.0 grams

Mass of water = 225 grams

Molar mass of KCl = 74.55 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: Total mass

Total mass = mass KCl + mass H2O

Total mass = 35.0 grams + 225.0 grams = 260.0 grams

Step 3: Calculate mass % of KCl

mass % KCl = (mass KCl/ total mass) * 100%

mass % KCl = (35.0 grams / 260.0 grams) *100%

mass % KCl = 13.5 %

Step 4: Calculate moles KCl

Moles KCl = mass KCl / molar mass

Moles KCl = 35.0 grams / 74.55 g/mol

Moles KCl = 0.469 moles

Step 5: Calculate moles H2O

Moles H2O = mass H2O / molar mass H2O

Moles H2O = 225.0 grams / 18.02 g/mol

Moles H2O = 12.49 moles

Step 6: Calculate total moles

Total moles = moles KCl + moles H2O

Total moles = 0.469 + 12.49 = 12.959

Step 7: Calculate mol fraction KCl

mol fraction = 0.469 / 12.959 = 0.0362

Step 8: Calculate molarity

C = moles KCl / volume

C = 0.469 moles / 0.239 L

C = 1.96 M

Step 9: Calculate molality KCl

Molality = moles KCl / mass solvent

Molality = 0.469 moles / 0.225 kg

Molality = 2.08 molal

Answer 2

A. The mass percent of KCl in the solution is 13.5%

B. The mole fraction of the ionic species KCl in the solution is 0.036

C. The molarity of the KCl in the solution is 1.967 M

D. The molality of the KCl in the solution is 2.09 M

A. Determination of the mass percent of KCl in the solution.

Mass of KCl = 35 g

Mass of water = 225 g

Mass of solution = 35 + 225

Mass of solution = 260 g

Percentage of KCl =?

$Percentage = \frac{mass of solute}{mass of solution} * 100 \\\\= \frac{35}{260} * 100$

= 13.5%

Therefore, the mass percent of KCl in the solution is 13.5%

B. Determination of the mole fraction of the ionic species KCl in the solution

• We'll begin by calculating the number of mole in 35 g of KCl.

Mass of KCl = 35 g

Molar mass of KCl = 39 + 35.5 = 74.5 g/mol

Mole of KCl =?

Mole = mass / molar mass

Mole of KCl = 35 / 74.5

Mole of KCl = 0.47 mole

• Next, we shall determine the number of mole in 225 g of water, H₂O

Mass of H₂O = 225 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mole of H₂O =?

Mole = mass / molar mass

Mole of H₂O = 225 / 18

Mole of H₂O = 12.5 mole

• Finally, we shall determine the mole fraction of the ionic species KCl.

Mole of KCl = 0.47 mole

Mole of H₂O = 12.5 mole

Total mole = 0.47 + 12.5 = 12.97 mole

Mole fraction of KCl =?

Mole fraction = mole of element / total mole

Mole fraction of KCl = 0.47 / 12.97

Mole fraction of KCl = 0.036

Therefore, the mole fraction of the ionic species KCl is 0.036

C. Determination of the molarity of KCl in the solution.

Mole of KCl = 0.47 mole

Volume = 239 mL = 239 / 1000 = 0.239 L

Molarity of KCl =?

Molarity = mole / Volume

Molarity of KCl = 0.47 / 0.239

Molarity of KCl = 1.967 M

Therefore, the molarity of KCl in the solution is 1.967 M

D. Determination of the molality of the KCl in the solution

Mole of KCl = 0.47 mole

Mass of water = 225 g = 225 / 1000 = 0.225 Kg

Molality of KCl =?

Molality = mole / mass of water (Kg)

Molality of KCl = 0.47 / 0.225

Molality of KCl = 2.09 M

Therefore, the molality of the KCl in the solution is 2.09 M

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