Molarity of $ \mathrm{H}_2\mathrm{SO}_4 $ = 0.5 M

Volume of $ \mathrm{H}_2\mathrm{SO}_4 $ = 50 mL = 0.050 L

Moles of $ \mathrm{H}_2\mathrm{SO}_4 = \text{Molarity} \times \text{Volume} = 0.5 \times 0.050 = 0.025 \, \text{mol} $

Determine the moles of $ \mathrm{NaOH} $ used for neutralizing the leftover acid:

Molarity of $ \mathrm{NaOH} $ = 1 M

Volume of $ \mathrm{NaOH} $ = 14.2 mL = 0.0142 L

Moles of $ \mathrm{NaOH} = 1 \times 0.0142 = 0.0142 \, \text{mol} $

Calculate moles of unreacted $ \mathrm{H}_2\mathrm{SO}_4 $:

Since it takes 2 moles of $ \mathrm{NaOH} $ to neutralize 1 mole of $ \mathrm{H}_2\mathrm{SO}_4 $:

$ \text{Moles of unreacted } \mathrm{H}_2\mathrm{SO}_4 = \frac{\text{Moles of } \mathrm{NaOH}}{2} = \frac{0.0142}{2} = 0.0071 \, \text{mol} $

Find the moles of $ \mathrm{H}_2\mathrm{SO}_4 $ that reacted with the metal:

$ \text{Moles of } \mathrm{H}_2\mathrm{SO}_4 \, (\text{reacted}) = \text{Initial moles} - \text{Unreacted moles} = 0.025 - 0.0071 = 0.0179 \, \text{mol} $

Use the reaction stoichiometry:

The equation for the reaction is:

$ \mathrm{M} + \mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{MSO}_4 + \mathrm{H}_2 $

From stoichiometry, 1 mole of metal (M) reacts with 1 mole of $ \mathrm{H}_2\mathrm{SO}_4 $. Therefore, moles of metal = moles of reacted $ \mathrm{H}_2\mathrm{SO}_4 = 0.0179 \, \text{mol} $.

Calculate the atomic weight of the metal:

$ \text{Atomic weight of metal} = \frac{\text{Mass of metal}}{\text{Moles of metal}} = \frac{0.43}{0.0179} \approx 24 \, \text{u} $

Thus, the atomic weight of the metal is approximately 24 u.

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