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The number of moles of KOH present in 200cm\(^3\) of 5.6 g dm\(^{-3}\) solution is ...

Chemistry
WAEC 2025

The number of moles of KOH present in 200cm\(^3\) of 5.6 g dm\(^{-3}\) solution is  [KOH = 56 g mol\(^{-1}\)]

  • A. 2.01
  • B. 1.12
  • C. 0.02
  • D. 0.01
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Correct Answer: Option C
Explanation

Given: 

Volume, V = 200 cm\(^3\)

Concentration, C = 5.6 g/dm\(^3\)

Molar mass of KOH, M = 56 g/mol

Number of moles, n = \(\frac{CV}{1000}\)

But because mass concentration was given in g/dm\(^3\) instead of molar concentration (mol/dm\(^3\)) to enable us calculate the number of moles, we'll have to convert concentration from mass to molar using the relationship below:

        Molar concentration (mol/dm\(^3\)) =  \(\frac{mass concentration}{Molar Mass}\)

                                                                = \(\frac{5.6}{56}\) = 0.1 mol/dm\(^3\)

Now, we can find the number of moles using, Number of moles, n = \(\frac{0.1 X  200}{1000}\) = 0.02 mole

The correct answer is option C

 


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