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  4. 25 mL of 3 M HCl were added to 75 mL of 0.05 M HCl. The molarity of HCl in the resulting solution is approximately

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Q. 25 mL of 3 M HCl were added to 75 mL of 0.05 M HCl. The molarity of HCl in the resulting solution is approximately

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Solution:

Molarity of resulting solution
$M_{Total} = \frac{M_1V_1}{V_{Total}} + \frac{M_2V_2}{V_{Total}}$
= $\frac{25 \times 3}{100} + \frac{75 \times 0.05}{100} $
= 0.75 + 0.0375 = 0.7875 M.