Problem 15 - A Complete Resource Book in Chemistry for JEE Main

Question

The hydrated salt \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{n} \mathrm{H}_{2} \mathrm{O}\) undergoes \(63 \%\) loss in mass on heating and becomes anhydrous. The value of \(\mathrm{n}\) is (a) 4 (b) 6 (c) 8 (d) 10

Step-by-Step Solution

Answer

The value of \( n \) is 10.

Step 1: Understand the Problem

We are given a hydrated salt of the form \( \text{Na}_{2} \text{CO}_{3} \cdot n \text{H}_2 \text{O} \) that loses \( 63\% \) of its mass when heated to become anhydrous. We need to find the value of \( n \), the number of water molecules per formula unit of the hydrated salt.

Step 2: Set Up the Equation

Assume the molar mass of the anhydrous salt \( \text{Na}_{2} \text{CO}_{3} \) is \( M_a = 106 \text{ g/mol} \) and the molar mass of water \( \text{H}_2\text{O} \) is \( M_w = 18 \text{ g/mol} \). The total molar mass of the hydrated salt is \( M_t = M_a + n \cdot M_w \).

Step 3: Express the Loss in Mass

A \( 63\% \) loss in mass due to the removal of water implies that \( 37\% \) of the original mass remains as the anhydrous salt. Therefore, we write the equation: \( 0.37 \, M_t = M_a \), where \( M_t \) is the initial molar mass including water.

Step 4: Solve for n

Substituting \( M_t = M_a + n \cdot M_w \) into the equation \( 0.37 \, M_t = M_a \), we get \( 0.37(M_a + n \cdot M_w) = M_a \). Simplify this: \( 0.37M_a + 0.37nM_w = M_a \). Rearrange to solve for \( n \): \( 0.37nM_w = M_a(1 - 0.37) \). Calculate \( n \): \( n = \frac{M_a(1 - 0.37)}{0.37M_w} = \frac{106 \times 0.63}{0.37 \times 18} \approx 10 \).

Step 5: Conclusion

The calculated value of \( n \) is close to \( 10 \), which matches one of the given options. Thus, the value of \( n \) indicating the number of water molecules in the hydrated salt is \( 10 \).

Key Concepts

Molar Mass Calculation

When working with chemicals, one crucial concept to understand is molar mass. It's the mass of one mole of a substance, typically expressed in grams per mole (g/mol). Calculating the molar mass involves adding up the atomic masses of all the atoms in a given formula. For instance, to find the molar mass of sodium carbonate (\(\text{Na}_{2} \text{CO}_{3}\)), you'll add:

• 2 sodium (Na) atoms
• 1 carbon (C) atom
• 3 oxygen (O) atoms

The atomic masses from the periodic table provide these values: Na = 23 g/mol, C = 12 g/mol, and O = 16 g/mol. Thus, the molar mass of \(\text{Na}_{2} \text{CO}_{3}\) is calculated as:\[2 \times 23 + 12 + 3 \times 16 = 106 \text{ g/mol}\]For compounds with more components, such as hydrated salts, additional water molecules must be included in the calculation. This step is crucial for accurately determining the compound's behavior and reactivity when subjected to various conditions, such as heating.

Anhydrous Compounds

Anhydrous compounds are substances that do not contain water molecules as part of their crystalline structure. In chemical reactions or processes, especially those involving heating, water molecules attached to compounds might be lost, leaving behind an anhydrous (water-free) form.

An example is sodium carbonate (\(\text{Na}_{2} \text{CO}_{3}\)), which, as a hydrated compound, includes water in its structure. Upon heating, the water evaporates, and the compound turns into its anhydrous form. This process is crucial in various industries where precise properties of a material are required, impacting factors like solubility and stability.

In the classroom or lab, when dealing with anhydrous compounds, students should be aware that the physical properties might vary significantly from their hydrated forms. This understanding helps when predicting behaviors during experiments or when making calculations related to storage and reactions.

Water of Crystallization

Water of crystallization refers to water molecules that are integrated into a crystal structure of a compound. These water molecules can drastically affect the physical and chemical properties of the compound.

For example, the hydrated salt sodium carbonate decahydrate (\(\text{Na}_{2} \text{CO}_{3} \cdot 10 \text{H}_2 \text{O}\)) contains ten water molecules for every unit of sodium carbonate. These water molecules are not just around but are part of the compound's crystalline matrix.

Understanding this concept is crucial, especially when performing experiments that involve heating hydrated salts. As these substances are heated, they undergo a transition to an anhydrous state when the water evaporates. This can affect the mass and structural integrity of the compound, making it a vital consideration in both laboratory settings and industrial applications. Knowing how to calculate and account for water of crystallization helps to predict these changes, ensuring accurate measurements and results.