Question

A mixture consisting of only sodium chloride (NaCl) and potassium chloride (KCl) weighs 1.0000g. When the mixture is dissolved in water and an excess of silver nitrate is added all the chloride ions associated with the original mixture are precipitated in soluble silver chloride (AgCl). The mass of silver chloride is formed to be 2.1476g. Calculate the mass percentage of sodium chloride and potassium chloride in the original mixture.

Short answer

The mass percentages of NaCl and KCl are 39.28% and 60.97%, respectively.

Step-by-step solution

Mass percentage

The formula of mass percentage is:

$\mathbf{Mass}\mathbf{\%}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solute}}{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{solution}}\mathbf{\times }\mathbf{100}\mathbf{\%}$

We are given the mixture consisting of sodium chloride (NaCl) and potassium chloride (KCl) and the mass of silver chloride to calculate the mass percentage of sodium chloride and potassium chloride involved.

 Step 2: Calculation of numbers of moles of sodium chloride and potassium chloride 

The mass of a mixture of sodium chloride and potassium chloride is 1.0000g.

Let us consider x the mass of KCL and y the mass of NaCl present in this mixture. Hence

x + y = 1.0000 …..(i)

The number of moles of sodium chloride and potassium chloride are calculated using their molar masses.

$MassofKCl=xg.KCl\times \frac{1mol.KCl}{74.55g.KCl}$

Calculation of the moles of AgCl

When the mixture is dissolved in water and excess of silver mixture is added, 2.1476g of silver chloride forms.

The number of moles of AgCl present in 2.1476g.

$$\begin{gathered} =2.1476g.AgCl\times \frac{1mol.AgCl}{143.32g} \\ =0.014985mol.AgCl \end{gathered}$$

From the molar formula of KCl, NaCl, and AgCl, it is clear that 1 mol of each of these compounds contains 1 mol Cl. As all the Cl in the 1.0000g mixture is to be precipitated as AgCl, the moles of Cl present in 2.1476g AgCl will be equal to the total moles of Cl present in the 1.0000g mixture.

Therefore:

$$\begin{gathered} \frac{\text{x}}{\text{74.55}}\text{mol +}\frac{\text{y}}{\text{58.44}}\text{mol = 0.014985mol} \\ \text{0.01341x + 0.01711y = 0.0014985 - - - - (2)} \end{gathered}$$

Solving the equation formed during the calculation of moles of the compound

Multiplying equation (1) with 0.01341 gives:

0.01341x + 0.01341y = 0.01341---(3)

Subtracting equation (3) from equation (2) gives

$$\begin{gathered} \text{0.01341x + 0.01711y = 0.0014985} \\ 0.01341x+0.01341y=0.01341 \\ -------------------- \\ 0.03052y=0.0119115 \\ y=0.39028 \end{gathered}$$

Subtracting the value of y in equation (1) gives:

$$\begin{gathered} \text{x + 0.39028 = 1.0000} \\ \text{x = 0.6097} \end{gathered}$$

Calculation of mass percentage of NaCl and KCl in the mixture

So, the masses of KCl and NaCl present in the mixture are 0.39028g KCl and 0.6097g NaCl.

Then the mass percentage of sodium and potassium chloride in the original mixture is its mass divided by the total mass of the original mixture multiplied by 100%.

Mass percentage of NaCl in the mixture =$$\begin{gathered} \frac{0.3928}{1g}\times 100\% \\ =39.28\% \end{gathered}$$

Mass percentage of KCl in the mixture =$$\begin{gathered} \frac{0.6097}{1g}\times 100\% \\ =60.97\% \end{gathered}$$

Hence, the mass percentages of NaCl and KCl are 39.28% and 60.97% respectively.