What volume of distilled water should be added to 400cm3 of 2.0 mole dm-3 \(H_{2}SO_{4}\) to obtain 0.20 mole dm-3 of solution>
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Correct Answer: Option D
Explanation:
\(400cm^{3} \to 2moldm^{-3}\)
\(0.4 dm^{3} \to 2moldm^{-3}\)
Multiplying in order to get the moles of the acid.
\(0.4dm^{3} \times 2moldm^{-3} = 0.8moles\)
After diluting to \(0.2moldm^{-3}\), the new volume will be
\(\frac{0.8}{0.2} = 4dm^{3}\)
Converting back to cm^3,
We have \(4 \times 1000cm^{3} = 4000cm^{3}\)
Hence, the volume of water added = \((4000 - 400)cm^{3} = 3600cm^{3}\)
\(400cm^{3} \to 2moldm^{-3}\)
\(0.4 dm^{3} \to 2moldm^{-3}\)
Multiplying in order to get the moles of the acid.
\(0.4dm^{3} \times 2moldm^{-3} = 0.8moles\)
After diluting to \(0.2moldm^{-3}\), the new volume will be
\(\frac{0.8}{0.2} = 4dm^{3}\)
Converting back to cm^3,
We have \(4 \times 1000cm^{3} = 4000cm^{3}\)
Hence, the volume of water added = \((4000 - 400)cm^{3} = 3600cm^{3}\)