Answer:
Molarity of solution is 1.10x10⁻³ M
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
Answer:
[tex]\large \boxed{\text{1.11 mol $\cdot$ dm}^{-3}}[/tex]
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
[tex]\text{ Mass of solution} = \text{1000 cm}^{3} \times \dfrac{\text{1.12 g solution}}{\text{1 cm}^{3}} = \text{1120 g solution}[/tex]
(b) Mass of NaOCl
[tex]\text{Mass of NaOCl} = \text{1120 g solution}\times \dfrac{\text{7.4 g NaOCl}}{\text{100 g solution}} = \text{82.9 g NaOCl}[/tex]
(c) Moles of NaOCl
[tex]\text{Moles of NaOCl} = \text{82.9 g NaOCl} \times \dfrac{\text{1 mol NaOCl}}{\text{74.44 g NaOCl}} = \text{1.11 mol NaOCl}[/tex]
(d) Molar concentration
[tex]\text{Molar concentration} = \ \dfrac{\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}}}$}[/tex]
