Question:medium

\(3\times10^{22}\) molecules of \(Na_2CO_3\) molecular weight \(=106\) present in \(500\text{ ml}\) of solution. The normality of the solution formed is \(\left(N=6\times10^{23}\text{ mol}^{-1}\right)\)

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First convert molecules into moles using Avogadro number. Then find molarity and multiply by valency factor to get normality.
  • \(0.1N\)
  • \(0.2N\)
  • \(0.4N\)
  • \(0.05N\)
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The Correct Option is B

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