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)

Assessment: WAEC SSCE - Chemistry - 2013 (Objective) Subject: Chemistry

Question 1 Report

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)

464.30 cm³

625.00 cm³

1000.00 cm³

2153.80 cm³

Answer Details

The molar mass of KOH (potassium hydroxide) is K + O + H = 39.0 + 16.0 + 1.00 = 56.0 g/mol. To find the amount of KOH needed to yield 6.5g, we divide 6.5 by the molar mass of KOH: 6.5g / 56.0 g/mol = 0.116 mol Since the concentration of KOH is given as 0.25 moldm^-3, we can use the formula: concentration (mol/dm³) = amount of substance (mol) / volume (dm³) to find the volume of the solution required: 0.25 mol/dm³ = 0.116 mol / volume (dm³) Solving for volume, we get: volume (dm³) = 0.116 mol / 0.25 mol/dm³ = 0.464 dm³ = 464.30 cm³ Therefore, the answer is 464.30 cm³.

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Stoichiometry And Chemical Reactions

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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)

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