Question

How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?

a. 45 mL of a 0.10 M solution:
mL

b. 330 mL of a 0.074 M solution:
mL

Answer 1

a. Approximately
`1.3\; \rm mL`.

b. Approximately
`7.2\; \rm mL`.

Step-by-step explanation:

The unit of concentration "
`\rm M`" is equivalent to "
`\rm mol \cdot L^(-1)`", which means "moles per liter."

However, the volume of both solutions were given in mililiters
`\rm mL`. Convert these volumes to liters:

`\displaystyle 45\; \rm mL = 45\; \rm mL * (1\; \rm L)/(1000\; \rm mL) = 0.045\; \rm L`.

`\displaystyle 330\; \rm mL = 330\; \rm mL * (1\; \rm L)/(1000\; \rm mL) = 0.330\; \rm L`.

In a solution of volume
`V` where the concentration of a solute is
`c`, there would be
`c \cdot V` (moles of) formula units of this solute.

Calculate the number of moles of
`\rm NaCl` formula units in each of the two solutions:

Solution in a.:

`n = c \cdot V = 0.045\; \rm L * 0.10\; \rm mol \cdot L^(-1) = 0.0045\; \rm mol`.

Solution in b.:

`n = c \cdot V = 0.330\; \rm L * 0.074\; \rm mol \cdot L^(-1) = 0.02442\; \rm mol`.

What volume of that
`3.4\; \rm M` (same as
`3.4 \; \rm mol \cdot L^(-1)`)
`\rm NaCl` solution would contain that many

For the solution in a.:

`\displaystyle V = (n)/(c) = (0.0045\; \rm mol)/(3.4\; \rm mol \cdot L^(-1)) \approx 0.0013\; \rm L`.

Convert the unit of that volume to milliliters:

`\displaystyle 0.0013\; \rm L = 0.0013\; \rm L * (1000\; \rm mL)/(1\; \rm L) = 1.3\; \rm mL`.

Similarly, for the solution in b.:

`\displaystyle V = (n)/(c) = (0.02442\; \rm mol)/(3.4\; \rm mol \cdot L^(-1)) \approx 0.0072\; \rm L`.

Convert the unit of that volume to milliliters:

`\displaystyle 0.0072\; \rm L = 0.0072\; \rm L * (1000\; \rm mL)/(1\; \rm L) = 7.2\; \rm mL`.