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

Question

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 (A) 0.055 M (B) 0.35 M (C) 0.787 M (D) 3 .05 M

Tardigrade Admin · Accepted Answer

Molarity of resulting solution MTotal = (M1V1/VTotal) + (M2V2/VTotal) = (25 × 3/100) + (75 × 0.05/100) = 0.75 + 0.0375 = 0.7875 M.

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

0.055 M

0.35 M

0.787 M

3 .05 M

Molarity of resulting solution
$M_{T o t a l} = \frac{M _{1} V _{1}}{V _{T o t a l}} + \frac{M _{2} V _{2}}{V _{T o t a l}}$
= $\frac{25 \times 3}{100} + \frac{75 \times 0.05}{100}$
= 0.75 + 0.0375 = 0.7875 M.

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

0.055 M

0.35 M

0.787 M

3 .05 M

Molarity of resulting solution MTotal​=VTotal​M1​V1​​+VTotal​M2​V2​​ = 10025×3​+10075×0.05​ = 0.75 + 0.0375 = 0.7875 M.

Molarity of resulting solution
$M_{T o t a l} = \frac{M _{1} V _{1}}{V _{T o t a l}} + \frac{M _{2} V _{2}}{V _{T o t a l}}$
= $\frac{25 \times 3}{100} + \frac{75 \times 0.05}{100}$
= 0.75 + 0.0375 = 0.7875 M.