To determine the normality (\(N\)) of the resulting solution, we need to consider the change in moles of the solute (\(H_2SO_4\)) after dilution.
Given:
Initial concentration of \(H_2SO_4\) = 0.5 M
Initial volume of solution = 1 L
Final volume of solution after dilution = 10 L
Normality (\(N\)) is defined as the number of gram equivalents of solute per liter of solution.
To calculate the normality, we need to determine the number of gram equivalents of \(H_2SO_4\) present in the solution.
The normality (\(N\)) can be calculated using the following formula:
\[N = \frac{{\text{{Number of moles of solute}} \times \text{{Number of equivalents per mole}}}}{{\text{{Volume of solution (in liters)}}}}\]
For \(H_2SO_4\), each mole of the solute is equivalent to 2 equivalents because it can donate two hydrogen ions (\(H^+\)).
First, let's calculate the initial number of moles of \(H_2SO_4\):
Number of moles of \(H_2SO_4\) = \(\text{{Initial concentration}} \times \text{{Initial volume}}\)
Number of moles of \(H_2SO_4\) = \(0.5, \text{M} \times 1 \, \text{L} = 0.5 \, \text{mol}\)
Since the dilution is performed by adding water, the number of moles of \(H_2SO_4\) remains constant. Therefore, after dilution, we still have 0.5 mol of \(H_2SO_4\).
Now, let's calculate the normality of the resulting solution:
Normality (\(N\)) = \(\frac{{\text{{Number of moles of }} H_2SO_4 \times \text{{Number of equivalents per mole}}}}{{\text{{Volume of solution}}}}\)
Number of equivalents per mole of \(H_2SO_4\) = 2 (since each mole of \(H_2SO_4\) is equivalent to 2 equivalents)
Normality (\(N\)) = \(\frac{{0.5 \, \text{mol} \times 2}}{{10 \, \text{L}}} = 0.1 \, \text{N}\)
Therefore, the normality of the resulting solution is 0.1 N.
Hence, the correct option is (4) 0.1 N. |
