Problem 62 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A mixture contains \(\mathrm{NaCl}\) and unknown chloride MCl. (i) \(1 \mathrm{~g}\) of this is dissolved in water. Excess of acidified \(\mathrm{AgNO}_{3}\) solution is added to it. \(2.567 \mathrm{~g}\) of white ppt. is formed. (ii) \(1 \mathrm{~g}\) of original mixture is heated to \(300^{\circ} \mathrm{C}\). Some vapours come out which are absorbed in acidified \(\mathrm{AgNO}_{3}\) solution, \(1.341 \mathrm{~g}\) of white precipitate was obtained. Find the molecular weight of unknown chloride. [1980]

Step-by-Step Solution

Answer

The molecular weight of the unknown chloride \( \mathrm{MCl} \) is approximately 53.5 g/mol.

Step 1: Understand the problem

You have a mixture containing \( \mathrm{NaCl} \) and an unknown chloride \( \mathrm{MCl} \). It reacts with acidified \( \mathrm{AgNO}_{3} \) to give a precipitate of \( \mathrm{AgCl} \). You need to find the molecular weight of \( \mathrm{MCl} \) based on given precipitate masses at two different conditions.

Step 2: Reaction understanding

When \( \mathrm{NaCl} \) or \( \mathrm{MCl} \) reacts with \( \mathrm{AgNO}_{3} \), \( \mathrm{AgCl} \) precipitate forms according to: \( \mathrm{NaCl (aq) + AgNO}_{3} \mathrm{(aq) \rightarrow AgCl (s) + NaNO}_{3} \mathrm{(aq)} \) and \( \mathrm{MCl (aq) + AgNO}_{3} \mathrm{(aq) \rightarrow AgCl (s) + MNO}_{3} \mathrm{(aq)} \). This means that the moles of chloride ions will be equal to the moles of \( \mathrm{AgCl} \) formed.

Step 3: Calculate moles of AgCl for full reaction

From condition (i), \( 2.567 \mathrm{~g} \) of \( \mathrm{AgCl} \) is formed. The molar mass of \( \mathrm{AgCl} \) is \( 143.3 \mathrm{~g/mol} \). Calculate the moles of \( \mathrm{AgCl} \): \( \frac{2.567}{143.3} \approx 0.0179 \mathrm{~mol} \).

Step 4: Analyze the heated mixture

On heating, chloride MCl decomposes and releases chloride ions as \( \mathrm{HCl(g)} \), which react to form \( 1.341 \mathrm{~g} \) of \( \mathrm{AgCl} \). Calculate \( moles \) of \( \mathrm{AgCl} \): \( \frac{1.341}{143.3} \approx 0.00936 \mathrm{~mol} \). This value corresponds to chloride ion solely from \( \mathrm{MCl} \).

Step 5: Determine moles of NaCl

Subtract the \( moles \) of chloride (from \( \mathrm{MCl} \)) from total \( moles \) of chloride: \( 0.0179 - 0.00936 = 0.00854 \mathrm{~mol} \). This is the \( moles \) of \( NaCl \).

Step 6: Calculate grams of NaCl and MCl

Grams of \( NaCl \): \( 0.00854 \times 58.5 = 0.499 \mathrm{~g} \). Since the total mixture weighs \( 1 \mathrm{~g} \), grams of \( \mathrm{MCl} \): \( 1 - 0.499 = 0.501 \mathrm{~g} \).

Step 7: Calculate molar mass of MCl

From condition (ii), we have \( 0.00936 \mathrm{~mol} \) of \( \mathrm{MCl} \), therefore: \( \mathrm{Molar\,\ mass\,\ of\,\ MCl = \frac{0.501}{0.00936} \approx 53.5\,\ g/mol} \).

Key Concepts

Chloride Reaction

In the world of chemistry, understanding how chloride ions participate in reactions is vital. When a chloride compound like sodium chloride (\( \mathrm{NaCl} \)) or an unknown chloride compound (\( \mathrm{MCl} \)) reacts with silver nitrate (\( \mathrm{AgNO}_{3} \)), a classic chloride reaction occurs. The result of this reaction is the formation of silver chloride (\( \mathrm{AgCl} \)), a white precipitate that suggests the presence of chloride ions. This type of reaction is represented by the following chemical equations:

• \( \mathrm{NaCl (aq) + AgNO}_{3} \mathrm{(aq) \rightarrow AgCl (s) + NaNO}_{3} \mathrm{(aq)} \)
• \( \mathrm{MCl (aq) + AgNO}_{3} \mathrm{(aq) \rightarrow AgCl (s) + MNO}_{3} \mathrm{(aq)} \)

These equations reveal that each mole of \( \mathrm{NaCl} \) or \( \mathrm{MCl} \) consumed produces one mole of \( \mathrm{AgCl} \) through this chloride reaction. This reaction underpins many analyses in chemistry, particularly in determining unknown chloride compounds and their quantities.

Precipitation Reaction

A precipitation reaction is one where reactants produce an insoluble product, known as a precipitate, in solution. When \( \mathrm{AgNO}_{3} \) is added to solutions containing chloride ions, \( \mathrm{AgCl} \) is precipitated out because it is insoluble in water. This is a key characteristic of precipitation reactions—they remove ions from the solution, causing clarity as the solid settles.
In our exercise, when the mixture containing \( \mathrm{NaCl} \) and \( \mathrm{MCl} \) is reacted with excess \( \mathrm{AgNO}_{3} \), the resulting white precipitate of \( \mathrm{AgCl} \) reveals the quantity of chloride ions originally present. By knowing the mass of the precipitate and the molar mass of \( \mathrm{AgCl} \), it's possible to calculate the moles of chloride ions. This illustrates how precipitation reactions are used to deduce information about reactant quantities and compositions.

Stoichiometry

Stoichiometry is the aspect of chemistry that deals with the quantitative relationships of reactants and products in chemical reactions. It involves calculations based on the balanced chemical equations to predict the amounts of substances consumed and produced. In our exercise, stoichiometry allows us to determine the molecular weight of the unknown chloride \( \mathrm{MCl} \).
Starting with the known mass of \( \mathrm{AgCl} \) precipitates formed, we calculate the moles of \( \mathrm{NaCl} \) and the unknown \( \mathrm{MCl} \) using their respective chemical equations. Here's what happens:

• The total moles of chloride ions precipitated from the first reaction are found from \( 2.567 \mathrm{~g} \) of \( \mathrm{AgCl} \).
• When heated, the unknown \( \mathrm{MCl} \) decomposes, evolving chloride ions as \( \mathrm{HCl(g)} \), which again react to form \( 1.341 \mathrm{~g} \) of \( \mathrm{AgCl} \).

By subtracting the moles from the second condition from the overall moles, we determine the moles of \( \mathrm{NaCl} \) and calculate its contribution to the mixture's mass. This systemic approach showcases stoichiometry's power in linking equations with real-world laboratory data to solve for unknowns.