Question

What is the molarity of the bleach solution of slide 28 (7.4% NaOCl by mass, density 1.12 g.cm-3)?

Answer 1

`\large \boxed{\text{1.11 mol $\cdot$ dm}^(-3)}`

Step-by-step explanation:

Molar concentration = moles/cubic decimetres

So, we need both the number of moles and the volume.

1. Volume

Assume a volume of 1 dm³.

That takes care of that.

2. Moles of NaOCl

(a) Mass of solution

`\text{ Mass of solution} = \text{1000 cm}^(3) * \frac{\text{1.12 g solution}}{\text{1 cm}^(3)} = \text{1120 g solution}`

(b) Mass of NaOCl

`\text{Mass of NaOCl} = \text{1120 g solution}* \frac{\text{7.4 g NaOCl}}{\text{100 g solution}} = \text{82.9 g NaOCl}`

(c) Moles of NaOCl

`\text{Moles of NaOCl} = \text{82.9 g NaOCl} * \frac{\text{1 mol NaOCl}}{\text{74.44 g NaOCl}} = \text{1.11 mol NaOCl}`

(d) Molar concentration

`\text{Molar concentration} = \ \frac{\text{1.11 mol NaOCl}}{\text{1 dm}^(3)} = \textbf{1.11 mol $\cdot$ dm}^{\mathbf{-3}}\\\\\text{The molar concentration of the NaOCl is $\large \boxed{\textbf{1.11 mol $\cdot$ dm}^{\mathbf{-3}}}$}`

Answer 2

Molarity of solution is 1.10x10⁻³ M

Step-by-step explanation:

Solute NaOCl

7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.

Molar mass of NaOCl = 74.45 g/m

Mol = Mass / Molar mass

7.4 g / 74.45 g/m = 0.099 moles

Density of solution = 1.12 g/mL

Density = Mass / volume

1.12g/mL = 100 g / volume

Volume = 100 g / 1.12g/mL = 89.3 mL

Molarity = mol /L

89.3 mL = 0.0893 L

0.099 moles / 0.0893 L = 1.10x10⁻³ M