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