Answer:
0.625 M
Explanation:
First, 1 cm³ = 1 mL. We can use the conversion to convert the cm³ into liters since molarity is just moles divided by volume(liters). We can use the molar mass of sodium carbonate and the total volume to figure out the molarity of the acid.
Solving:
[tex]\begin{itemize}\\\item Given:\\ \item Mass of \(\text{Na}_2\text{CO}_3\) sample: \(26.50 \, \text{g}\) \item Volume of solution: \(500.00 \, \text{cm}^3\) \item Volume of \(\text{Na}_2\text{CO}_3\) solution used: \(25.00 \, \text{cm}^3\) \item Volume of \(\text{H}_2\text{SO}_4\) solution used: \(20.00 \, \text{cm}^3\)\end{itemize}[/tex]
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[tex]\subsection*{Moles of \(\text{Na}_2\text{CO}_3\)~:}\[\text{Molar mass of } \text{Na}_2\text{CO}_3 = 2(\text{Na}) + \text{C} + 3(\text{O}) \]\[= 2(22.99) + 12.01 + 3(16.00) \]\[= 45.98 + 12.01 + 48.00 \]\[= 105.99 \, \text{g/mol}\]\[\text{Moles of } \text{Na}_2\text{CO}_3 = \frac{26.50 \, \text{g}}{105.99 \, \text{g/mol}} \]\[\boxed{= 0.250 \, \text{mol}}\][/tex]
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[tex]\subsection*{Molarity of \(\text{Na}_2\text{CO}_3\) Solution:}\[\text{Molarity of } \text{Na}_2\text{CO}_3 = \frac{0.250 \, \text{mol}}{0.500 \, \text{L}} \]\[\boxed{= 0.500 \, \text{M}}\][/tex]
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[tex]\subsection*{ Moles of \(\text{Na}_2\text{CO}_3\) in 25.00 cm\(^3\) of Solution:}\[\text{Volume} = 25.00 \, \text{cm}^3 = 0.025 \, \text{L}\]\[\text{Moles of } \text{Na}_2\text{CO}_3 \text{ in } 25.00 \, \text{cm}^3 = 0.500 \, \text{M} \times 0.025 \, \text{L} \]\[\boxed{= 0.0125 \, \text{mol}}\][/tex]
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[tex]\subsection*{Balanced Chemical Equation:}\[\text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2\]From the balanced equation, 1 mole of \(\text{Na}_2\text{CO}_3\) reacts with 1 mole of \(\text{H}_2\text{SO}_4\).\[\text{} 0.0125 \text{ moles of } \text{Na}_2\text{CO}_3 \text{ will react with } 0.0125 \text{ moles of } \text{H}_2\text{SO}_4 \text{}.\][/tex]
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[tex]\subsection*{Molarity of \(\text{H}_2\text{SO}_4\):}\[\text{Volume} = 20.00 \, \text{cm}^3 = 0.020 \, \text{L}\]\[\text{Molarity of } \text{H}_2\text{SO}_4 = \frac{0.0125 \, \text{mol}}{0.020 \, \text{L}} \]\[\boxed{= 0.625 \, \text{M}}\][/tex]
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Therefore, the molarity of the H2SO4 acid is 0.625.
