Answer:
a. Approximately [tex]1.3\; \rm mL[/tex].
b. Approximately [tex]7.2\; \rm mL[/tex].
Explanation:
The unit of concentration "[tex]\rm M[/tex]" is equivalent to "[tex]\rm mol \cdot L^{-1}[/tex]", which means "moles per liter."
However, the volume of both solutions were given in mililiters [tex]\rm mL[/tex]. Convert these volumes to liters:
[tex]\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L[/tex].
[tex]\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L[/tex].
In a solution of volume [tex]V[/tex] where the concentration of a solute is [tex]c[/tex], there would be [tex]c \cdot V[/tex] (moles of) formula units of this solute.
Calculate the number of moles of [tex]\rm NaCl[/tex] formula units in each of the two solutions:
Solution in a.:
[tex]n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol[/tex].
Solution in b.:
[tex]n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol[/tex].
What volume of that [tex]3.4\; \rm M[/tex] (same as [tex]3.4 \; \rm mol \cdot L^{-1}[/tex]) [tex]\rm NaCl[/tex] solution would contain that many
For the solution in a.:
[tex]\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L[/tex].
Convert the unit of that volume to milliliters:
[tex]\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL[/tex].
Similarly, for the solution in b.:
[tex]\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L[/tex].
Convert the unit of that volume to milliliters:
[tex]\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL[/tex].
