Problem 157 - A Complete Resource Book in Chemistry for JEE Main

Question

On heating \(1.763 \mathrm{~g}\) of hydrated \(\mathrm{BaCl}_{2}\) to dryness, \(1.505 \mathrm{~g}\) of anhydrous salt remained. Number of moles of \(\mathrm{H}_{2} \mathrm{O}\) present in one mole of the hydrated \(\mathrm{BaCl}_{2}\) is [Mol. wt. of anhydrous \(\mathrm{BaCl}_{2}\) is 208\(]\).

Step-by-Step Solution

Answer

Two moles of \(\mathrm{H_2O}\) are present per mole of hydrated \(\mathrm{BaCl}_2\).

Step 1: Calculate the Mass of Water Lost

The mass of the hydrated compound is \(1.763\, \text{g}\), and the mass of the anhydrous \(\mathrm{BaCl}_2\) remaining after heating is \(1.505\, \text{g}\). Therefore, the mass of water lost is the difference: \[1.763\, \text{g} - 1.505\, \text{g} = 0.258\, \text{g}.\]

Step 2: Calculate Moles of Anhydrous \(\mathrm{BaCl}_2\)

Using the molecular weight of anhydrous \(\mathrm{BaCl}_2\), which is \(208\, \text{g/mol}\), calculate the moles of \(\mathrm{BaCl}_2\):\[\text{Moles of } \mathrm{BaCl}_2 = \frac{1.505\, \text{g}}{208\, \text{g/mol}} \approx 0.00723\, \text{mol}.\]

Step 3: Calculate Moles of Water

Using the molar mass of water \(18\, \text{g/mol}\), calculate the moles of water lost:\[\text{Moles of } \mathrm{H_2O} = \frac{0.258\, \text{g}}{18\, \text{g/mol}} \approx 0.01433\, \text{mol}.\]

Step 4: Determine the Hydrate Formula

The number of moles of water per mole of \(\mathrm{BaCl}_2\) is given by the ratio:\[\text{Number of moles of } \mathrm{H_2O} \text{ per mole of } \mathrm{BaCl}_2 = \frac{0.01433\, \text{mol}}{0.00723\, \text{mol}} \approx 1.98.\]This is approximately 2 moles of \(\mathrm{H_2O}\) per mole of \(\mathrm{BaCl}_2\).

Step 5: Confirm the Hydrate Formula

Based on the calculations, the hydrated compound is \(\mathrm{BaCl}_2 \cdot 2\mathrm{H_2O}\), indicating that each mole of hydrated \(\mathrm{BaCl}_2\) contains approximately 2 moles of water.

Key Concepts

Hydrated Compounds

Hydrated compounds are substances that include water molecules that are integrated into their solid structure. These water molecules are not random; they are fixed into the crystalline framework and play a role in determining the compound's properties. The term "hydrated" denotes the presence of water.
In the formula representation, water molecules are typically shown separately from the main compound, such as in \( \text{BaCl}_2 \cdot 2\text{H}_2\text{O} \). This simply means that for every molecule of \( \text{BaCl}_2 \), there are 2 water molecules.
Hydrated compounds are interesting because heating them often removes the water, turning them into "anhydrous" forms. The weight difference before and after heating helps determine the amount of water present in the original hydrated form.

Mole Concept

The mole is a fundamental concept in chemistry that provides a bridge between the atomic or molecular scale and the macroscopic scale of materials you can handle. Essentially, one mole represents \( 6.022 \times 10^{23} \) entities, whether they are atoms, molecules, ions, or other particles.
The mole allows chemists to count particles by weighing them. So, instead of saying "I have \( 120.22 \times 10^{23} \) molecules of water," we say "I have 20 moles of water." The concept is crucial for converting between the mass of a substance and the number of moles, helping simplify and streamline stoichiometric calculations.

• It ties together mass (grams) and quantity (number of molecules).
• It is pivotal in calculating chemical formulas and reactions.

Understanding this concept is key in analyzing reactions and for calculating quantities needed for reactions or those produced.

Chemical Formulas

Chemical formulas are shorthand ways to represent the composition of molecules and compounds. They indicate the types and numbers of atoms present in a molecule.
For instance, in our compound \( \text{BaCl}_2 \cdot 2\text{H}_2\text{O} \), the formula tells us that each unit consists of one barium ion, two chloride ions, and two water molecules.
These notations provide a compact method to convey a wealth of information about a chemical compound. It tells us:

• The constituent elements.
• The proportions of these elements within the compound.

Chemical formulas are essential for stoichiometry because they help balance chemical reactions and guide us in the synthesis and decomposition of chemical compounds.

Molar Mass Calculations

Molar mass is a critical parameter in chemistry that provides the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It gives an idea of how much one mole of particles weighs.
To find the molar mass, one needs to add up the atomic masses of all atoms in the molecular formula of the compound. For example, the molar mass of anhydrous \( \text{BaCl}_2 \) is \( 208 \, \text{g/mol} \).
This calculation is useful because it links the atomic scale with the real-world scale, allowing you to determine how much of a substance is needed for a reaction:

• It converts between grams and moles, which is pivotal for quantitative chemistry.
• It is crucial in determining the proportions needed for reactions.

Understanding how to calculate and use molar mass effectively is vital for any student engaging in chemical experiments and reactions.