Question

Quantity of $Na_2CO_3$ required to prepare 500 ml of 0.1N $Na_2CO_3$ is _______________.

• A. 2.65
• B. 0.265
• C. 26.5
• D. 0.00265

Hint:

For a $1\text{N}\ Na_2CO_3$ solution, you need $53\text{ g/L}$. For a $0.1\text{N}$ solution, you need $5.3\text{ g/L}$. Therefore, for half a liter ($500\text{ ml}$), you need exactly half of that weight: $2.65\text{ g}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To prepare a solution of a specific normality ($N$), we must calculate the required mass of the solute based on its equivalent weight.
Normality is defined as the number of gram equivalents of solute dissolved per liter of solution.

Step 2: Key Formula or Approach:
1. The equivalent weight of a compound is:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{Valency factor } (n)} \]
2. The mass of solute ($W$) required is calculated as:
\[ W = \text{Normality } (N) \times \text{Equivalent Weight} \times \text{Volume in Liters } (V) \]

Step 3: Detailed Explanation:
Let us perform the calculations step-by-step:
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Step 1: Find the molar mass of Sodium Carbonate ($Na_2CO_3$):
\[ \text{Molar Mass} = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106\text{ g/mol} \]
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Step 2: Determine the valency factor ($n$) of $Na_2CO_3$:
Since $Na_2CO_3$ is a salt that dissociates into $2\ Na^+$ and $CO_3^{2-}$, the total positive or negative charge is $2$. Thus, $n = 2$.
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Step 3: Calculate the equivalent weight of $Na_2CO_3$:
\[ \text{Equivalent Weight} = \frac{106}{2} = 53\text{ g/eq} \]
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Step 4: Calculate the mass required to prepare $500\text{ ml}$ ($0.5\text{ L}$) of $0.1\text{N}$ solution:
\[ W = N \times \text{Equivalent Weight} \times V \]
\[ W = 0.1 \times 53 \times 0.5 \]
\[ W = 5.3 \times 0.5 = 2.65\text{ grams} \]
Therefore, $2.65\text{ g}$ of $Na_2CO_3$ is required, matching Option (A).

Step 4: Final Answer:
The quantity of $Na_2CO_3$ required is 2.65.