Question

\[1.0g\] of a sample containing \[NaCl\] , \[KCl\] and some other impurity is dissolved in excess of water and treated with excess of \[AgN{O_3}\] solution. A \[2.0g\] precipitate of \[AgCl\] separates out the sample is \[23\% \] by mass in sodium. Determine the mass percentage of \[KCl\] in the sample.
A) \[45\% \]
B) \[29.28\% \]
C) \[10\% \]
D) \[52.10\% \]

Answer 1

Hint: Here we need to determine the mass percentage. Mass percent of an element in a compound can be calculated by dividing mass of element in one mole of a compound by compound molar mass. Then it can be multiplied by \[100\]. Actually the mass percent formula is expressed as solving to get the molar mass of every element in one mole of a compound.

Complete step-by-step answer:
Here the mass of sample given is \[1.0g\]
The percentage of sample by mass in sodium is \[23\% \]
Mass of sodium here can be calculated as \[\dfrac{{23}}{{100}} \times 1.0 = 0.23g\]
Here atomic masses of \[Na\] and \[Cl\] are \[23.0g/mol\] and \[35.5g/mol\] respectively.
When we take the case of \[NaCl\] , \[0.23g\] of \[Na\] can be corresponded to \[\dfrac{{35.5}}{{23.0}} \times 0.23 = 0.355g\] of chlorine.
We know the molar mass of \[AgCl\] is \[143.5g/mol\] and that the atomic mass of \[Cl\] is \[35.5g/mol\] .
\[2.0g\]\[AgCl\] of \[AgCl\] precipitate is separated out as given in the question.
Therefore here \[2.0g\] of \[AgCl\] thus corresponds to \[\dfrac{{35.5}}{{143.5}} \times 2.0 = 0.495\] grams of chlorine.
Out of the total \[0.495g\] of chlorine \[0.355g\] of chlorine is from \[NaCl\].
Therefore the remaining that is , \[0.495 - 0.355 = 0.140\]gram is from \[KCl\] .
We know the molar mass of \[KCl\] is \[74.6g/mol\].
Atomic mass of chlorine is \[35.5g/mol\]
Then \[0.140g\] of chlorine corresponds to \[\dfrac{{74.6}}{{35.5}} \times 0.140 = 0.2928g\] of \[KCl\].
The mass of the sample is \[1.0g\] as given in question. The mass of \[KCl\] is \[0.2928g\] as calculated above.
Therefore mass percent of \[KCl\] in sample can be calculated as the following,
\[\dfrac{{0.2928}}{{1.0}} \times 100 = 29.28\% \]

Hence option B is the correct answer.
Note: Mass percent is a way of expressing degree. More than that , it describes components involved in a particular mixture. The solution composition is actually described in mass percentage. It deals with showing mass of solute which is present during a given mass of solution. Here the number of moles is expressed in mass percent by moles.