Chapter 14.2, Problem 26PP

Interpretation Introduction

Interpretation:

Amount of HCl in solution is to be calculated.

Concept introduction:

Dilution is the process of decreasing the concentration of a stock solution by adding more solvent to the solution. The solvent added is usually the universal solvent, known as water. The more solvent you add, the more diluted the solution will get.

In dilution, the amount of solute does not change, the number of moles is the same before and after dilution.

If subscript "1" represents initial and "2" represents the final values of the quantities involved, we have:

${\mathrm{n}}_1 =M_1\times V_1 and n_2=M_2\times V_2$

Here, molarity is used as a unit of concentration and n₁ and n₂ are the number of moles before and after dilution respectively.

n₁  = n₂   and thus,

The equation for dilution is

$\begin{array}{c}{M}_1{V}_1=M_2{V}_2\\ Stocksolution=Dilutedsolution\end{array}$

Where

M₁ = molarity of the stock solution

V₁ = volume of stock solution

M₂ = molarity of the diluted solution

V₂ = volume of diluted solution

Answer to Problem 26PP

91.15 gmHCl is present in the solution.

As we know, in dilution, the amount of solute does not change, the number of moles is the same before and after dilution.

Here, the equation is requiredn₁ = M₁V₁

M₁ = molarity of the stock solution V₁ = volume of stock solution

Data Given: M₁ = molarity of the stock solution = 5.0 M

V₁ = volume of stock HCl solution = 0.5 L

n₁ = M₁V₁

= 5.0 M × 0.5 L = 2.5 mol Number of moles of HCl $=\frac{MassofHCl}{MolecularweightofHCl}$

Calculate the mass of HCl from above equation:

$\begin{array}{l}MassofHCl=NumberofmolesofHCl\times MolecularweightofHCl\hfill \\ =2.5mol\times 36.46gm/mol\hfill \\ =91.15gm\hfill \end{array}$

Thus, 91.15 gmHCl is present in the solution.

Explanation of Solution

As we know, in dilution, the amount of solute does not change, the number of moles is the same before and after dilution.

Here, the equation is requiredn₁ = M₁V₁

M₁ = molarity of the stock solution V₁ = volume of stock solution

Data Given: M₁ = molarity of the stock solution = 5.0 M

V₁ = volume of stock HCl solution = 0.5 L

n₁ = M₁V₁

= 5.0 M × 0.5 L = 2.5 mol Number of moles of HCl $=\frac{MassofHCl}{MolecularweightofHCl}$

Calculate the mass of HCl from above equation:

$\begin{array}{l}MassofHCl=NumberofmolesofHCl\times MolecularweightofHCl\hfill \\ =2.5mol\times 36.46gm/mol\hfill \\ =91.15gm\hfill \end{array}$

Thus, 91.15 gmHCl is present in the solution.