Question

1.An aqueous solution of 0.975 M hydrochloric acid, HCl, has a density of 1.02 g/mL. The percent by mass of HCl in the solution is ___ %.
2.An aqueous solution of 1.29 M ethanol, CH3CH2OH, has a density of 0.988 g/mL. The percent by mass of CH3CH2OH in the solution is ___%.
3.An aqueous solution is 40.0% by mass silver nitrate, AgNO3, and has a density of 1.47 g/mL. The molarity of silver nitrate in the solution is ___M.

Answer 1

Answer:

The percent mass of HCl in the solution is 3.48  %

The percent mass of ethanol in the solution is 6.01  %

The molarity of silver nitrate in the solution is 3.46 M

Explanation:

• Percent by mass = mass of solute/100 g of solution

Density always referrs to solution data.

1.02 g/mL = 100 g / volume of solution

Volume of solution = 98.04 mL

Molarity = moles of solute in 1L of solution (or mmoles of solute in 1mL)

M . 98.04 mL = mmoles of solute → 0.975 M . 98.04 mL = 95.59 mmoles

mmoles . PM (mg / mmol) = mg → 95.59 mmol . 36.45 mg/mmol = 3484 mg.

We convert data to g → 3484 mg . 1g/ 1000 mg = 3.48 g

The percent mass of HCl in the solution is 3.48  %

• Percent by mass = mass of solute/100 g of solution

Density always referrs to solution data.

0.988 g/mL = 100 g / volume of solution

Volume of solution = 101.21 mL

Molarity = moles of solute in 1L of solution (or mmoles of solute in 1mL)

M . 101.21 mL = mmoles of solute → 1.29 M . 101.21 mL = 130.6 mmoles

mmoles . PM (mg / mmol) = mg → 130.6 mmol . 46.07 mg/mmol = 6015 mg.

We convert data to g → 6015 mg . 1g/ 1000 mg = 6.01 g

The percent mass of ethanol in the solution is 6.01  %

• Percent by mass = mass of solute/100 g of solution

Density always referrs to solution data.

1.47 g/mL = 100 g / volume of solution

Volume of solution = 68.03 mL

Molarity = moles of solute in 1L of solution (or mmoles of solute in 1mL)

40 g = 40000 mg

mg / PM (mg / mmol) = mmol → 40000 mg / 169.87 mg/mmol

= 235.4 mmoles

mmoles / mL = Molarity  → 235.4 mmol / 68.03 mL = 3.46 M