Problem 8 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A sample of a hydrate of barium chloride weighing \(61 \mathrm{~g}\) was heated until all the water of hydration is removed. The dried sample weighed \(52 \mathrm{~g}\). The formula of the hydrated salt is : (atomic mass, \(\mathrm{Ba}=137\) \(\mathrm{amu}, \mathrm{Cl}=35.5 \mathrm{amu}\) ) [Main Online April 10,2015] (a) \(\mathrm{BaCl}_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{BaCl}_{2} .3 \mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{BaCl}_{2} \cdot \mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The formula of the hydrated salt is \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\) (option d).

Step 1: Calculate the mass of water lost

First, subtract the mass of the dehydrated sample from the mass of the hydrated sample to find the mass of the water that was lost during heating. This is given as:\[\text{Mass of water lost} = 61\,\text{g} - 52\,\text{g} = 9\,\text{g}\]

Step 2: Calculate the moles of anhydrous barium chloride

Next, calculate the moles of the anhydrous \(\text{BaCl}_2\) using its molar mass. The formula mass of \(\text{BaCl}_2\) is calculated as:\[137\,\text{amu (Ba)} + 2 \times 35.5\,\text{amu (Cl)} = 208\,\text{g/mol}\]The moles of \(\text{BaCl}_2\) are then given by:\[\text{Moles of BaCl}_2 = \frac{52\,\text{g}}{208\,\text{g/mol}} \approx 0.25\,\text{mol}\]

Step 3: Calculate the moles of water lost

The molar mass of water (\(\text{H}_2\text{O}\)) is 18 g/mol. The moles of water are calculated as:\[\text{Moles of water} = \frac{9\,\text{g}}{18\,\text{g/mol}} = 0.5\,\text{mol}\]

Step 4: Determine the molar ratio

Now, find the ratio of moles of water to moles of \(\text{BaCl}_2\). This can be calculated as:\[\text{Ratio} = \frac{\text{Moles of water}}{\text{Moles of } \text{BaCl}_2} = \frac{0.5}{0.25} = 2\]This means there are 2 moles of water for every mole of \(\text{BaCl}_2\).

Step 5: Identify the formula for the hydrated salt

Based on the ratio from Step 4, the formula for the hydrated barium chloride is \(\text{BaCl}_2 \cdot 2\text{H}_2\text{O}\), which corresponds to option (d).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry focused on the quantitative relationships between reactants and products in a chemical reaction. It provides the tools to understand how much of each substance is involved in a reaction, allowing us to predict yields and optimize processes. In the case of hydrate determination, stoichiometry helps us relate the mass of the hydrate, the anhydrous compound remaining after heating, and the water that is removed. By understanding these relationships, we can establish how many moles of water are associated with each mole of the anhydrous compound.

When faced with such exercises, it is important to approach them strategically:

• First, calculate the mass loss due to heating, which equates to the mass of water initially present in the hydrate form.
• Second, determine the amount of the anhydrous compound by considering its leftover mass.
• Use stoichiometric principles to relate these mass changes to molar quantities.

Through this systematic approach, stoichiometry provides clarity and precision in chemical calculations.

Mole Calculation

Mole calculations are a core part of chemistry, allowing us to connect macroscopic amounts of substances to their microscopic quantities. A mole, in chemical terms, is an amount that links the number of particles (atoms, molecules, etc.) to a quantifiable mass. For water of hydration calculations, it's crucial to know both how to count moles and to convert between moles and grams using substances' molar masses.

• To calculate moles, divide the mass of the substance by its molar mass. For example, for barium chloride with a molar mass of 208 g/mol, 52 g of it would contain about 0.25 mol of barium chloride.
• Similarly, for the water lost, 9 g of water corresponds to 0.5 mol when divided by the water's molar mass of 18 g/mol.

Understanding mole calculations underpins countless chemical reactions and analyses, including determining the hydration number in hydrates.

Molar Mass Concept

The concept of molar mass is extremely helpful when you need to convert between grams and moles of a substance. Molar mass is defined as the mass of one mole of a particular substance, with units of g/mol. Each element contributes to the molar mass of a compound based on its atomic mass and its quantity within the molecule.

For barium chloride, the molar mass is calculated by adding:

• The atomic mass of barium (137 amu) with
• Two times the chlorine atomic mass (2 x 35.5 amu)

This results in a molar mass of 208 g/mol for barium chloride. Again, for water, which often serves as a straightforward example, the molar mass is calculated as:

• Two hydrogen atoms (2 x 1 amu) plus one oxygen atom (16 amu)

Summing to 18 g/mol.

Mastery of molar mass is critical for mole-to-mass and mass-to-mole conversions, which are instrumental when solving chemical problems.

Hydration Number Calculation

The hydration number in a hydrate is essential as it tells us how many water molecules are associated with each molecule of the compound. To determine this number, a step-by-step approach is adopted.

• First, calculate the moles of the anhydrous compound and the moles of evaporated water.
• Then, determine the ratio of moles of water to moles of the compound by dividing the two values.

For instance, in the original exercise, with 0.5 moles of water and 0.25 moles of barium chloride, the ratio is calculated as 0.5/0.25, resulting in a hydration number of 2. This indicates two water molecules are attached to each mole of barium chloride.

Understanding hydration number calculations is crucial when dealing with hydrates as it allows chemists to accurately determine chemical formulas and comprehend the structural characteristics of the compounds involved.