Calculate the number of moles of solute present in each of the following solutions.

Part A 255mL of 1.70M HNO3 (aq)

Part B 45.0mg of an aqueous solution that is 1.35m NaCl. Assume that for dilute aqueous solutions, the mass of the solvent is the mass of solution.

Part C 80.0g of an aqueous solution that is 1.50% sucrose (C12H22O11) by mass

A. 0.433 moles of HNO₃

B. 6.07*10⁻⁵ moles of NaCl

C. 3.51*10⁻³ moles of sucrose

Explanation:

Part A.

Molarity. volume = moles

1.70 mol/L. 0.255L = 0.433 moles

Part B.

1.35 m of NaCl means, a molal concentration (m). Moles of solute in 1kg of solvent.

Let's convert 1 kg to mg for the rule of three

1 kg = 1*10⁶ mg

So in 1*10⁶ mg of solvent, we have 1.35 moles of solute (NaCl)

In 45 mg, we would have (45. 1.35) / 1*10⁶ = 6.07*10⁻⁵ moles of NaCl

Part C. 1.50 % by mass means, 1.50 g of solute, in 100g of solution.

Let's make a rule of three:

In 100 g of solution we have 1.50 g of sucrose

In 80 g of solution, we would have (80. 1.50) / 100 = 1.2 g of sucrose.

Then, let's convert the mass in moles (mass / molar mass)

1.2 g / 342 g/m = 3.51*10⁻³ moles