Question:
The mass of $Na_{2}CO_{3}$ (in g) (M.wt=106) present in 1.0 L of 0.05 M solution is
The mass of $Na_{2}CO_{3}$ (in g) (M.wt=106) present in 1.0 L of 0.05 M solution is
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Mass = Molarity $\times$ Molecular Weight for a 1-liter solution.
Updated On: Jul 4, 2026
- 0.53
- 53.0
- 26.5
- 5.30
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Molarity ($M$) is defined as the number of moles of solute per liter of solution. The formula is $M = \frac{mass}{M.wt \times Volume(L)}$.
Step 2: Meaning
A 0.05 M solution contains 0.05 moles of $Na_{2}CO_{3}$ in every 1 liter of the solution.
Step 3: Analysis
We are given $M = 0.05$, $M.wt = 106$, and $V = 1.0$ L. Rearranging the formula to find mass: $mass = M \times M.wt \times V = 0.05 \times 106 \times 1.0$.
Step 4: Conclusion
Calculation gives $mass = 5.30$ g. Final Answer: (D)
Molarity ($M$) is defined as the number of moles of solute per liter of solution. The formula is $M = \frac{mass}{M.wt \times Volume(L)}$.
Step 2: Meaning
A 0.05 M solution contains 0.05 moles of $Na_{2}CO_{3}$ in every 1 liter of the solution.
Step 3: Analysis
We are given $M = 0.05$, $M.wt = 106$, and $V = 1.0$ L. Rearranging the formula to find mass: $mass = M \times M.wt \times V = 0.05 \times 106 \times 1.0$.
Step 4: Conclusion
Calculation gives $mass = 5.30$ g. Final Answer: (D)
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