Question

An impure sample of sodium chloride that weighed \[0.5g\] gave \[0.9g\] of silver chloride as precipitate on treatment with excess silver nitrate solution. Calculate the percentage purity of the sample:
A.\[72%\]
B.\[78%\]
C.\[80%\]
D.None of the above

Answer 1

Hint:Percentage purity of a sample can be calculated by taking the ratio of the pure substance present in the solution, and the total weight of the sample, then multiplying it by \[100\] gives the percentage.
-Sodium chloride and silver nitrate, upon reaction, forms silver chloride which is a white precipitate. The sample whose percentage purity we have to determine is sodium chloride

Complete step by step answer:
-The given reaction in the question can be chemically represented as
$NaCl+AgNO3\to AgCl+NaNO3$
-Here, as we can see, one mole of sodium chloride reacts with one mole of silver nitrate in order to give one mole of silver chloride and one mole of sodium nitrate. This reaction can generally be characterised as a precipitation reaction, as precipitate of \[AgCl\] is formed. This method of precipitation of silver chloride by using silver nitrate as a reactant, is also used in the process of gravimetric analysis in the field of analytical chemistry. This method determines the presence of concentration of chloride ions in the given unknown solution, as it forms precipitate by forming a bond with the silver ion.
-In order to determine the percentage purity of the sample, we will first calculate the amount of reactant involved in reaction in grams.
\[143.5g\] of \[AgCl\] is obtained from \[58.5g\] of pure \[NaCl\], as we already know one mole of each reactant is involved and weight of one mole of \[AgCl\] and \[NaCl\] is \[143.5g\] and \[58.5g\] respectively.
-Now the question says we have to take 0.9g of \[AgCl\], as precipitate,
So, $\dfrac{58.5\times 0.9}{143.5}g=0.36g$
We got, \[0.36g\] of pure \[NaCl\] which was obtained by \[0.9g\] of \[AgCl\].
-In order to calculate the percentage purity, we need to divide the amount of pure \[NaCl\] with the total mass of \[NaCl\] given in the question and then multiply it with \[100\], to get the percentage.
$\dfrac{0.36\times 100}{0.5}=72%$
-We take the ratios of value of pure sodium chloride and the total weight of sodium chloride which is given in the answer, and the percentage purity came out to be \[72%\], so the appropriate answer would be option A.
So the correct answer is option A.

Note:
-One mole of any substance contains the same amount of that substance which is equal to the atomic weight of that substance.
-The percentage purity of sodium chloride is the amount of pure sodium chloride present in the total amount of sodium chloride.