Question

calculate the molarity of a 750.0 mL solution prepared by dissolving 65.00 g of zinc acetate into water

Answer 1

Answer on Question #49815 – Chemistry - Inorganic Chemistry

Calculate the molarity of a 750.0 mL solution prepared by dissolving 65.00 g of zinc acetate into water

Solution:

$$V_{\text{sol}} = 750.0 \text{ mL} = 0.75 \text{ L}$$ $$m(\text{Zn(CH}_3\text{COO)}_2) = 65.00 \text{ g}$$ $$c(\text{Zn(CH}_3\text{COO)}_2) - ?$$

In chemistry, the molar concentration, it is also called molarity, $c_i$ is defined as the amount of a constituent $n_i$ (usually measured in moles – hence the name) divided by the volume of the mixture $V_{\text{sol}}$:

$$c_i = \frac{n_i}{V_{\text{sol}}}; \quad n_i = \frac{m_i}{Mw_i} \quad \text{so} \quad c_i = \frac{m_i}{Mw_i \times V_{\text{sol}}}.$$

Molecular mass or molecular weight refers to the mass of a molecule. It is calculated as the sum of the mass of each constituent atom multiplied by the number of atoms of that element in the molecular formula:

$$Mw = \sum A_i$$

In case of zinc acetate:

$$\begin{array}{l} Mw(\text{Zn(CH}_3\text{COO)}_2) = A(\text{Zn}) + 4 \times A(\text{C}) + 6 \times A(\text{H}) + 4 \times A(\text{O}) \\ = 65.38 + 4 \times 12.01 + 6 \times 1.01 + 4 \times 15.99 = 183.44 \text{ g/mol} \end{array}$$ $$c(\text{Zn(CH}_3\text{COO)}_2) = \frac{65.00 \text{ g}}{183.44 \text{ g/mol} \times 0.75 \text{ L}} = 0.47 \text{ mol/L}$$

Answer: 0.47 mol/L.

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