Question

You dissolve a 1.28-g sample of nacl in a total volume of 125.0 ml of solution. your lab partner has 1.50 m aqueous nacl and wants to make a solution with the same concentration and volume as yours. how much of the solution does your lab partner need to use?

Answer 1

Answer:

The 14.59 mL of solution is needed by our lab partner.

Explanation:

Mass of the NaCl = 1.28 g

Moles of NaCL =$n=\frac{1.28 g}{58.5 g/mol}=0.02190 mol$

$Molarity=\frac{n}{V(L)}$

V = Volume of the solution in Liters.

$V_1$= 125.0 mL = 0.1250 L

Molarity of our solution = $M_1$

$M_1=\frac{0.02190 mol}{0.1250 L}=0.1750 M$

Molarity of the lab partner's solution = $M_2=1.50 M$

Volume of the lab partner's solution = $V_2$ = ?

$M_1V_1=M_2V_2$

$V_2=\frac{0.1750 M\times 0.1250 L}{1.50 M}$

$V_2=0.01459 L=14.59 mL$

The 14.59 mL of solution is needed by our lab partner.