Question

in a schools laboratory, students require 50.0 ml of 2.50 m h₂so₄ 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

Explanation:

Step1: Identify known values

We know the formula for dilution is \( M_iV_i = M_fV_f \), where \( M_i = 18.0\ M \) (initial molarity), \( M_f = 2.50\ M \) (final molarity), and \( V_f = 50.0\ mL \) (final volume). We need to find \( V_i \) (initial volume).

Step2: Rearrange the formula to solve for \( V_i \)

From \( M_iV_i = M_fV_f \), we can rearrange it to \( V_i=\frac{M_fV_f}{M_i} \).

Step3: Substitute the known values into the formula

Substitute \( M_f = 2.50\ M \), \( V_f = 50.0\ mL \), and \( M_i = 18.0\ M \) into the formula:
\( V_i=\frac{2.50\ M\times50.0\ mL}{18.0\ M} \)
First, calculate the numerator: \( 2.50\ M\times50.0\ mL = 125\ mmol \) (since \( M\times V \) in mL gives mmol for dilute solutions, but here we are just doing the arithmetic). Then divide by the denominator: \( \frac{125}{18.0}\approx6.94\ mL \).

Answer:

6.94 mL