Question

You need to make 150.0 mL of a 0.10 M $\mathrm{NaCl}$solution. You have solid $\mathrm{NaCl}$and your lab partner has a 2.5 M $\mathrm{NaCl}$solution. Explain how you each make the 0.10 M $\mathrm{NaCl}$ solution.

Short answer

We can prepare 0.10 M solution by diluting $6.0\mathrm{mL}$of 2.5 M$\mathrm{NaCl}$ solution to 150.0 m.

Step-by-step solution

Number of moles

The mass of a material made up of an equal number of fundamental units is defined as a mole. The number of chemical elements is expressed as a mole fraction. The value of$6.023x{10}^{23}$ is equivalent to one mole of any material (Avogadro number). It can be used to quantify the end products of a chemical process. Mol is the unit of measurement.

The formula for calculating the amount of moles is as follows:

Number of Moles$=\frac{\mathrm{Mass}\mathrm{of}\mathrm{substance}}{\mathrm{Mass}\mathrm{of}\mathrm{one}\mathrm{mole}}$

Determination of mass of 1.5×10-2mol NaCl

No. of moles $\mathrm{NaCl}$of in 150.0mL of 0.10 M $\mathrm{NaCl}$

$=150.0\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\times 0.10\mathrm{M}$

$=150.0\mathrm{mL}\times \frac{1\mathrm{L}}{1000\mathrm{mL}}\times 0.10\frac{\mathrm{mol}}{\mathrm{L}}$

$=1.5\times {10}^{-2}\mathrm{mol}$

Here the molar mass of $\mathrm{NaCl}=58.44\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$

Thus, the mass of $=1.5\times {10}^{-2}\mathrm{mol}$$\mathrm{NaCl}=1.5\times {10}^{-2}\mathrm{mol}\mathrm{mol}\mathrm{NaCl}\times \frac{{\displaystyle 58.44\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.}}{{\displaystyle 1\mathrm{mol}\mathrm{NaCl}}}$

$=0.88\mathrm{gNaCl}$

0.10 M$\mathrm{NaCl}$ solution may be made by dissolving 0.88 g solid $\mathrm{NaCl}$in 150.0 mL water.

Determination of volume of desired NaCl

Here the concentration of $\mathrm{NaCl}$solution in stock${\mathrm{M}}_1=2.5\mathrm{M}$

The Volume of$\mathrm{NaCl}$solution in stock required${\text{=V}}_1$

The concentration of$\mathrm{NaCl}$solution in stock desired, ${\mathrm{M}}_2=0.10\mathrm{M}$

The Volume of $\mathrm{NaCl}$solution desired ${\mathrm{V}}_2=150.0\mathrm{mL}$

By dilution Principle, ${\mathrm{V}}_1{\mathrm{M}}_1={\mathrm{V}}_2{\mathrm{M}}_2$

$150.0\mathrm{mL}\times 0.10\mathrm{M}={\mathrm{V}}_2\times 2.5\mathrm{M}$

${\mathrm{V}}_2=\frac{150.0\mathrm{mL}\times 0.10\mathrm{M}}{2.5\mathrm{M}}$

role="math" localid="1650353844007" ${\mathrm{V}}_2=6.0\mathrm{mL}$

Therefore, $6.0\mathrm{mL}$of $2.5\mathrm{M}$$\mathrm{NaCl}$