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
Explanation
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Molar mass KOH = 39.1 + 16 + 1 = 56.1 g/mol
Mass of 0.25 mol KOH = 56.1 g/mol *0.25 mol = 14.025 g
1.0 dm³ 0.25 M KOH solution contains 14.025 g KOH
Volume that contains 6.5 g KOH = 6.5 g / 14.025 g/dm³ =0.463 dm³ or 463 mL.

The correct answer is A. 464.30 cm3.
To solve this problem, we need to use the following formula:
V = n/M
where:
* V is the volume of the solution
* n is the number of moles of KOH
* M is the molarity of the solution
We know that the molarity of the solution is 0.25 moldm-3 and that the mass of KOH is 6.5g. We can use the molar mass of KOH (56.1 g/mol) to calculate the number of moles of KOH:
n = \frac{m}{M} = \frac{6.5 \{ g}}{56.1 \{ g/mol}} = 0.116 \{ mol}
Now we can plug in the values of n and M into the formula to calculate the volume of the solution:
V = \frac{n}{M} = \frac{0.116 \{ mol}}{0.25 \{ moldm-3}} = 0.464 \{ dm-3} = 464.30 \{ cm3}
Therefore, the volume of 0.25 moldm-3 solution of KOH that would yield 6.5g of of solid KOH on evaporation is 464.30 cm3.


