Chapter 14, Problem 74A

How much $CaC{l}_2$ , in grams, is needed to make 2.0 L ofa 3.5M solution?

Interpretation Introduction

Interpretation:

The amount of CaCl₂have 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 74A

770.7 gm of CaCl₂ is needed to make 2.0 L of a 3.5 M solution.

Explanation of Solution

Given information:

Molarity of CaCl₂ solution = 3.5 M

Volume of CaCl₂ solution = 2 L

Write the expression for Molality.

$\mathrm{Molarity}ofthesolution=\frac{Numberofmolesof the solute}{ Volume of the solution(inlitre)}$

Calculate the number of moles of $CaC{l}_2$.

$\begin{array}{l}Numberofmolesof CaC{l}_2=Volume of CaC{l}_2 solution\times MolarityofCaC{l}_{2}solution\\ Numberofmolesof CaC{l}_2=2L\times 3.5mol/L=7mole\end{array}$

Calculate the mass of $CaC{l}_2$.

$\begin{array}{l}NumberofmolesofCaC{l}_2=\frac{MassofCaC{l}_2}{MolecularweightofCaC{l}_2}\\ MassofCaC{l}_2=NumberofmolesofCaC{l}_2\times MolecularweightofCaC{l}_2\\ MassofCaC{l}_2=7mol\times 110.1gm/mol=770.7gm\end{array}$

(Note: Molecular weight of CaCl₂ = 110.1 gm/mol)

Thus, the mass of CaCl₂ = 770.7 gm