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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

User E H
by
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1 Answer

3 votes

Answer:

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.

User Sua Morales
by
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