The volume of 0.25 moldm-3 solution of KOH that would yield 6.5g of of solid KOH on evaporation is (K = 39.0; o = 16.0; H = 1.00)
number of moles = molar concentration x volume(\(dm^3\))
molar concentration(C) = 0.25mol/dm3, volume = ? and number of mole(n) = \(\frac{mass}{molar mass}\) = \(\frac{6.5}{56}\) ( where 56g/mol = molar mass of KOH)
\(\frac{6.5}{56}\) = 0.25 x V
V = \(\frac{0.11607}{0.25}\) = 0.46428dm^3 → 464.3cm^3
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