Question

1. if 25.0 ml of 1.50 m hydrochloric acid is diluted to a volume of 500.0 ml, by the addition of water, what is the molarity of the dilute acid? a. 0.0750 m b. 0.0230 m c. 0.2350 m d. 0.0110 m e. 0.5020 m

Explanation:

Step1: Recall the dilution formula

\(M_1V_1 = M_2V_2\), where \(M_1\) is the initial molarity, \(V_1\) is the initial volume, \(M_2\) is the final molarity, and \(V_2\) is the final volume.

Step2: Substitute the given values into the formula

Given \(M_1=1.50\space M\), \(V_1 = 25.0\space mL\), \(V_2=500.0\space mL\). We need to find \(M_2\).
From \(M_1V_1 = M_2V_2\), we can solve for \(M_2\) as \(M_2=\frac{M_1V_1}{V_2}\).
Substitute the values: \(M_2=\frac{1.50\space M\times25.0\space mL}{500.0\space mL}\)

$$M_2=\frac{1.50\times25.0}{500.0}\space M=\frac{37.5}{500.0}\space M = 0.0750\space M$$

Answer:

A. \(0.0750\space M\)