Question

if 6.32 ml of a 5.79 m solution of hydrochloric acid, hcl, is diluted to a volume of 200.0ml by the addition of water, what is the concentration of the dilute solution?
5.47 m
0.183 m
0.199 m
1.06 m
6.67 m

Explanation:

Step1: Identify the dilution formula

The formula for dilution is $M_1V_1 = M_2V_2$, where $M_1$ is the initial - concentration, $V_1$ is the initial volume, $M_2$ is the final concentration, and $V_2$ is the final volume.

Step2: Assign the given values

We are given that $M_1 = 5.79\ M$, $V_1=6.32\ mL$, and $V_2 = 200.0\ mL$. We need to find $M_2$.

Step3: Rearrange the formula to solve for $M_2$

From $M_1V_1 = M_2V_2$, we can solve for $M_2$ by the formula $M_2=\frac{M_1V_1}{V_2}$.

Step4: Substitute the values into the formula

$M_2=\frac{5.79\ M\times6.32\ mL}{200.0\ mL}$.
First, calculate $5.79\times6.32 = 36.6928$. Then, $\frac{36.6928}{200.0}=0.183464\ M\approx0.183\ M$.

Answer:

$0.183\ M$