Answer:
The 14.59 mL of solution is needed by our lab partner.
Explanation:
Mass of the NaCl = 1.28 g
Moles of NaCL =[tex]n=\frac{1.28 g}{58.5 g/mol}=0.02190 mol[/tex]
[tex]Molarity=\frac{n}{V(L)}[/tex]
V = Volume of the solution in Liters.
[tex]V_1[/tex]= 125.0 mL = 0.1250 L
Molarity of our solution = [tex]M_1[/tex]
[tex]M_1=\frac{0.02190 mol}{0.1250 L}=0.1750 M[/tex]
Molarity of the lab partner's solution = [tex]M_2=1.50 M[/tex]
Volume of the lab partner's solution = [tex]V_2[/tex] = ?
[tex]M_1V_1=M_2V_2[/tex]
[tex]V_2=\frac{0.1750 M\times 0.1250 L}{1.50 M}[/tex]
[tex]V_2=0.01459 L=14.59 mL[/tex]
The 14.59 mL of solution is needed by our lab partner.
