Problem 68 - A Complete Resource Book in Chemistry for JEE Main

Question

The percentage weight of \(\mathrm{Zn}\) in white vitriol \(\left[\mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\right]\) is approximately equal to \((\mathrm{Zn}=65\) \(\mathrm{S}=32, \mathrm{O}=16\) and \(\mathrm{H}=1\) ) (a) \(21.56 \%\) (b) \(32.58 \%\) (c) \(22.65 \%\) (d) \(26.55 \%\)

Step-by-Step Solution

Answer

(c) 22.65%

Step 1: Determine Molar Mass of Each Component

First, calculate the molar mass of each element present in white vitriol \( \mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} \).- \( \mathrm{Zn} = 65 \, \text{g/mol} \)- \( \mathrm{S} = 32 \, \text{g/mol} \)- \( \mathrm{O} = 16 \, \text{g/mol} \)- \( \mathrm{H} = 1 \, \text{g/mol} \)

Step 2: Calculate Molar Mass of White Vitriol

Use the elemental masses to find the molar mass of white vitriol:\[ \mathrm{ZnSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O} = \mathrm{Zn} + \mathrm{S} + 4(\mathrm{O}) + 14(\mathrm{H}) + 7(\mathrm{O}) \]Calculate each part:- \( \mathrm{Zn} = 65 \, \text{g/mol} \)- \( \mathrm{S} = 32 \, \text{g/mol} \)- \( 4(\mathrm{O}) = 4 \times 16 = 64 \, \text{g/mol} \)- \( 7(\mathrm{H}_{2}\mathrm{O}) = 7 \times (2 \times 1 + 16) = 126 \, \text{g/mol} \)Adding these, the molar mass of white vitriol = \( 65 + 32 + 64 + 126 = 287 \, \text{g/mol} \).

Step 3: Calculate the Percentage Weight of Zn

Now, determine how much of the molar mass is due to zinc by dividing the molar mass of \( \mathrm{Zn} \) by the total molar mass, then multiply by 100 to find the percentage weight:\[ \frac{65}{287} \times 100 \approx 22.65\% \].

Step 4: Select the Correct Answer

The calculated percentage weight of \( \mathrm{Zn} \) in white vitriol is approximately \( 22.65\% \). Thus, the correct choice is \((c)\).

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is crucial in chemistry. It represents the mass of a mole of a substance and is usually measured in grams per mole (g/mol). To calculate the molar mass of a compound, you follow these basic steps:

• Identify all the elements present in the chemical formula.
• Determine the number of atoms of each element in the compound.
• Multiply the atomic mass of each element by the number of atoms of that element in the formula.
• Sum all the values to get the molar mass of the compound.

For example, in white vitriol, \( \mathrm{ZnSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O} \), the molar mass is calculated by adding the contributions of each element: zinc, sulfur, oxygen, and hydrogen. Follow the detailed steps as demonstrated above to ensure you understand this foundational concept.

White Vitriol

White vitriol is a common name for zinc sulfate heptahydrate, with a chemical formula of \( \mathrm{ZnSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O} \). This compound is known for its crystalline structure and is used in various applications. In the context of chemistry exercises, understanding white vitriol involves identifying the molecules and ions that make up the compound:

• Zinc (\( \mathrm{Zn} \)) acts as the central metallic element.
• Sulfur (\( \mathrm{S} \)) is bound to oxygen to form the sulfate (\( \mathrm{SO}_4 \)) ion.
• Water molecules are part of its crystalline form as water of crystallization.

This composition makes white vitriol an interesting subject of study, particularly when analyzing its chemical reactions and uses in industry and pharmacology.

Zinc Sulfate

Zinc sulfate is a compound that combines zinc, sulfur, and oxygen. It's typically encountered as zinc sulfate heptahydrate, \( \mathrm{ZnSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O} \), which means seven water molecules are associated with each formula unit. Zinc sulfate is an essential material in a wide range of industries, including:

• Fertilizer production: Zinc is a micronutrient vital for plant growth.
• Pharmaceuticals: It’s used in dietary supplements to prevent zinc deficiency.
• Metal plating: Helps in corrosion protection and as a mordant in dyeing.

Understanding zinc sulfate's basics and its molar mass is essential for those studying chemistry. It’s not only significant in industrial applications but also plays a role in educational exercises like the calculation of percentage composition.