Calculate the mass of ethanoic acid.
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's the solution for question 2:
Question 2: $15.0 \text{cm}^3$ of ethanoic acid ($\text{CH}_3\text{COOH}$) was dissolved in water to make $500 \text{cm}^3$ of solution. Calculate the concentration of the solution in moles per litre. [C=12, H=1, O=16, density of ethanoic acid is $1.05 \text{g/cm}^3$]
Step 1: Calculate the mass of ethanoic acid.
Given:
Volume of ethanoic acid = $15.0 \text{ cm}^3$
Density of ethanoic acid = $1.05 \text{ g/cm}^3$
Mass = Density $\times$ Volume
$$ \text{Mass of CH}_3\text{COOH} = 1.05 \text{ g/cm}^3 \times 15.0 \text{ cm}^3 = 15.75 \text{ g} $$
Step 2: Calculate the molar mass of ethanoic acid ($\text{CH}_3\text{COOH}$).
Atomic masses: C = 12, H = 1, O = 16
Formula: $\text{CH}_3\text{COOH}$ (which is $\text{C}_2\text{H}_4\text{O}_2$)
$$ \text{Molar mass of CH}_3\text{COOH} = (2 \times 12) + (4 \times 1) + (2 \times 16) $$
$$ \text{Molar mass of CH}_3\text{COOH} = 24 + 4 + 32 = 60 \text{ g/mol} $$
Step 3: Calculate the moles of ethanoic acid.
$$ \text{Moles of CH}_3\text{COOH} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{15.75 \text{ g}}{60 \text{ g/mol}} = 0.2625 \text{ mol} $$
Step 4: Calculate the concentration of the solution in moles per litre.
Given:
Moles of ethanoic acid = $0.2625 \text{ mol}$
Total volume of solution = $500 \text{ cm}^3 = 0.500 \text{ L}$
Concentration = $\frac{\text{Moles}}{\text{Volume (L)}}$
$$ \text{Concentration} = \frac{0.2625 \text{ mol}}{0.500 \text{ L}} = 0.525 \text{ mol/L} $$
The concentration of the solution is $\boxed{\text{0.525 mol/L}}$.
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