**Passage 2 (continued):** <br> **Q2:** The weight of in 1 L of the mixture is: — Redox Reactions and Volumetric Analysis Chemistry Question
Question
**Passage 2 (continued):** <br> **Q2:** The weight of $Na_2CO_3$ in 1 L of the mixture is:
Answer: B
💡 Solution & Explanation
**Step 1:** From previous logic, $Na_2CO_3$ in $25$ mL reacts with $30$ mL of $0.1 M$ $HCl$. <br>**Step 2:** Meq in $25$ mL = $30 \times 0.1 = 3$. <br>**Step 3:** Meq in $1000$ mL = $3 \times (1000/25) = 120$ meq. <br>**Step 4:** Mass = $(\text{Meq} \times \text{Eq. Wt}) / 1000 = (120 \times 53) / 1000 = 6.36$ g. (Note: Calculation check: $120 \times 106 / 2 / 1000 = 6.36$).
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