Question

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

Answer 1

Answer:

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