Question

4.1: solution concentration/water as a solvent
isotonic saline is a 0.15 m aqueous solution of nacl. how would you prepare 0.80 l of isotonic saline from a 6.0 m stock solution?

Explanation:

Step1: Use dilution formula

$M_1V_1 = M_2V_2$, 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.
We know that $M_1 = 6.0\ M$, $M_2=0.15\ M$, and $V_2 = 0.80\ L$.

Step2: Solve for $V_1$

Rearranging the formula $V_1=\frac{M_2V_2}{M_1}$.
Substitute the values: $V_1=\frac{0.15\ M\times0.80\ L}{6.0\ M}$.
$V_1=\frac{0.12\ mol}{6.0\ M}=0.02\ L$.
Then add $(0.80 - 0.02)=0.78\ L$ of water to $0.02\ L$ of the $6.0\ M$ stock solution to get $0.80\ L$ of $0.15\ M$ solution.

Answer:

Take $0.02\ L$ of the $6.0\ M$ stock solution and add $0.78\ L$ of water.