Chapter 15, Problem 37A

Interpretation Introduction

Interpretation: The mass of calcium carbonate precipitated needs to be determined.

Concept Introduction:

The molarity is the number of moles of the solute dissolved per liter volume of the solution. It is represented in mathematical term such that,

  $\begin{array}{l}M=\text{molarity=}\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{volume}\text{of}\text{the}\text{solution}}\\ M=\frac{n}{V}\end{array}$

Where, n is the number of moles,

V is the volume of the solution.

Answer to Problem 37A

The mass of the calcium carbonate which gets precipitate is $\text{0}\text{.559}\text{g}$ .

Explanation of Solution

Given information:

Volume of the calcium chloride is, $V_1=36.\text{2mL}$

Volume of the sodium carbonate, $V_2=3\text{7}\text{.7mL}$

Molarities of calcium chloride and sodium carbonate are $\text{0}\text{.158M}$ and $0.14\text{9M}$ respectively.

Molar mass of calcium carbonate is $100.1\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.$

Calculation:

The number of moles is given as,

  $\begin{array}{l}M=\frac{n}{V}\\ \Rightarrow n=MV\end{array}$

By substituting the given values in the above formula and then the result is,

The number of moles of calcium chloride is,

  $\begin{array}{c}n=MV\\ =\text{36}{\text{.2×10}}^{\text{-3}}\text{L×}\left(\text{0}\text{.158}\raisebox{1ex}{$\text{mol}$}\!\left/ \!\raisebox{-1ex}{$\text{L}$}\right.\right)\\ \text{=5}{\text{.719×10}}^{\text{-3}}\text{mol}\end{array}$

Similarly, the number of moles of sodium carbonate,

  $\begin{array}{c}n=MV\\ =37.5\times 10^{-3}L\times \left(0.149\raisebox{1ex}{$mol$}\!\left/ \!\raisebox{-1ex}{$L$}\right.\right)\\ =5.587\times 10^{-3}mol\end{array}$

The number of moles is always conserved during the dilution. Therefore, the number of moles of calcium carbonate is $5.587\times {10}^{-3}\text{mol}$ .

So, the mass of the calcium carbonate is,

  $\begin{array}{l}n=\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{Þmass=}\text{n×}\text{molar}\text{mass}\\ \text{=}\text{5}{\text{.587×10}}^{\text{-3}}\text{×100}\text{.1}\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.\\ \text{=}\text{0}\text{.559}\text{g}\end{array}$

Conclusion

The mass of calcium carbonate is $0.\text{559}\text{g}$ .