Chapter 14, Problem 110A

How many grams of calcium nitrate $(Ca{(N{O}_3)}_2)$ wouldyou need to prepare 3.00 L of a 0.500M solution?

Interpretation Introduction

Interpretation:

Amount of calcium nitratehave to be calculated.

Concept introduction:

Molarity: Molarity is defined as the number of moles of solute present in one litre of the solution In expression, $\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

$\mathrm{Number}ofmolesofsolute=\frac{Massofthesolute}{Molecularweightofthesolute}$

Answer to Problem 110A

246.14 gm of calcium nitrate $\mathrm{Ca}{\left(N{O}_3\right)}_2$ is needed to prepare 3.0 L of a 0.5M solution.

Explanation of Solution

Given information:

Volume of the solution = 3.0 L

Molarity of the $\mathrm{Ca}{\left(N{O}_3\right)}_2$ solution = 0.5 M.

Write the expression for Molality of the solution.

$Molarityofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

$\begin{array}{l}Numberofmolesof Ca{\left(N{O}_3\right)}_2=Molarityofthesolution\times Volume of the solution(inlitre)\\ =0.5M\times 3.0L=1.5mole\end{array}$

$\begin{array}{l}MassofCa{\left(N{O}_3\right)}_2=NumberofmolesofCa{\left(N{O}_3\right)}_2\times MolecularweightofCa{\left(N{O}_3\right)}_2\\ =1.5mole\times 164.09gm/mole=246.14gm\end{array}$

(Note: Molecular weight of $\mathrm{Ca}{\left(N{O}_3\right)}_2$ = 164.09 gm/mol)

Mass of calcium nitrate = 246.14 gm.