Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

\(4 \mathrm{~g}\) caustic soda is dissolved in \(100 \mathrm{cc}\) of solution. The normality of solution is (a) 1 (b) \(0.8\) (c) \(0.6\) (d) \(0.10\)

Step-by-Step Solution

Answer

The normality of the solution is (a) 1.

Step 1: Identify the Molecular Weight of Caustic Soda

Caustic soda, also known as sodium hydroxide (NaOH), has a molecular weight of \(23 + 16 + 1 = 40\) g/mol.

Step 2: Determine the Equivalent Weight

Normality involves equivalent weight. For NaOH, it is a strong base, the equivalent weight is the same as the molecular weight, which is \(40\) g/equivalent.

Step 3: Calculate the Number of Equivalents

Using the formula \( \text{equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}} \), we find that the number of equivalents of NaOH is \( \frac{4}{40} = 0.1 \) equivalents.

Step 4: Convert Solution Volume to Liters

Since normality is expressed in equivalents per liter, convert the solution's volume from cc to liters: \(100 \text{ cc} = 0.1 \text{ L}.\)

Step 5: Calculate the Normality

Normality is calculated as \(\text{Normality} = \frac{\text{equivalents}}{\text{volume in liters}} = \frac{0.1}{0.1} = 1\).

Key Concepts

Molecular Weight of Caustic Soda

When dealing with chemical solutions, understanding the concept of molecular weight can be incredibly helpful. Caustic soda, better known in the scientific community as sodium hydroxide (NaOH), has a specific molecular structure. This compound comprises one sodium (Na) atom, one oxygen (O) atom, and one hydrogen (H) atom. To find the molecular weight, we sum the atomic weights of these components:

• Sodium (Na): 23 g/mol
• Oxygen (O): 16 g/mol
• Hydrogen (H): 1 g/mol

This gives a total molecular weight of 23 + 16 + 1 = 40 g/mol. This number is crucial as it breaks down how much one mole of this compound weighs, providing a foundational value for further calculations in chemistry.

Equivalent Weight Calculation

The concept of equivalent weight is essential when determining the normality of a solution. Equivalent weight is a term used to describe the amount of a substance that reacts with or replaces one mole of hydrogen ions. In the case of strong bases like NaOH, the molecular weight and equivalent weight are identical. This is because NaOH disassociates completely and provides one hydroxide ion (OH-) per molecule in solution. Hence, for sodium hydroxide, the equivalent weight equals 40 g/equivalent, making calculations straightforward since no extra conversion is needed in this case. This similarity simplifies our calculations and is a crucial step in understanding how much of a substance will react in a given solution.

Converting Volume Units

Many students stumble upon the step of converting measurements, but it can be simple with a bit of practice. Chemical reactions in solution are often measured concerning liters. Therefore, it's vital to accurately convert volumes to the appropriate units for consistency in calculations.

• In the metric system, 1 liter equals 1000 cc (cubic centimeters).
• Thus, to convert 100 cc to liters, you divide by 1000.
• This means that 100 cc is equivalent to 0.1 liters.

This conversion is critical in the formula for normality, which requires liters, not cubic centimeters, to accurately determine the concentration of solutions in equivalent terms.

Calculating Equivalents

Equivalents are fundamental in understanding how a solution behaves chemically. To calculate equivalents, you use the formula:\[ \text{Equivalents} = \frac{\text{mass (g)}}{\text{equivalent weight (g/equiv)}}\]For NaOH, with a mass of 4 grams and an equivalent weight of 40 g/equivalent, the calculation would be:\[ \frac{4}{40} = 0.1 \text{ equivalents}\]This calculation provides a quantitative figure representing the amount of reactive substance present in the solution. Knowing how to calculate equivalents can empower us to determine the concentration of a solution more deeply and contribute to our broader understanding of chemical reactions. It is a useful skill for accurately solving chemistry problems related to solution concentration.