QuestionAugust 27, 2025

In a school's laboratory students require 50.0 mL of 2.50 M H_(2)SO_(4) for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use M_(i)V_(i)=M_(f)V_(f) 0.900 mL 1.11 mL 6.94 mL 7.20 mL

In a school's laboratory students require 50.0 mL of 2.50 M H_(2)SO_(4) for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use M_(i)V_(i)=M_(f)V_(f) 0.900 mL 1.11 mL 6.94 mL 7.20 mL
In a school's laboratory students require 50.0 mL of 2.50 M H_(2)SO_(4) for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use M_(i)V_(i)=M_(f)V_(f) 0.900 mL 1.11 mL 6.94 mL 7.20 mL

Solution
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Answer

6.94 mL Explanation 1. Identify known values M_i = 18.0 \, \text{M} (initial concentration), V_i = ? (initial volume), M_f = 2.50 \, \text{M} (final concentration), V_f = 50.0 \, \text{mL} (final volume). 2. Apply dilution formula Use the formula M_iV_i = M_fV_f. Substitute known values: 18.0 \times V_i = 2.50 \times 50.0. 3. Solve for V_i Rearrange to find V_i: V_i = \frac{2.50 \times 50.0}{18.0}. 4. Calculate V_i V_i = \frac{125.0}{18.0} = 6.94 \, \text{mL}.

Explanation

1. Identify known values<br /> $M_i = 18.0 \, \text{M}$ (initial concentration), $V_i = ?$ (initial volume), $M_f = 2.50 \, \text{M}$ (final concentration), $V_f = 50.0 \, \text{mL}$ (final volume).<br /><br />2. Apply dilution formula<br /> Use the formula $M_iV_i = M_fV_f$. Substitute known values: $18.0 \times V_i = 2.50 \times 50.0$.<br /><br />3. Solve for $V_i$<br /> Rearrange to find $V_i$: $V_i = \frac{2.50 \times 50.0}{18.0}$.<br /><br />4. Calculate $V_i$<br /> $V_i = \frac{125.0}{18.0} = 6.94 \, \text{mL}$.
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