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------------------------------------------------ How many milliliters of a 0.330 M KOH solution contain [tex]1.69 \times 10^{-2} \text{ mol}[/tex] of KOH?

Answer :

51.2 mL of the 0.330 mm KOH solution contains 1.69×10−2 mol of KOH. we can use the formula:

Volume = (amount of solute) / (molarity)

To calculate the volume of the solution containing a given number of moles of solute, we can use the formula:

Volume = (amount of solute) / (molarity)

Here, the molarity of the KOH solution is 0.330 mm or 0.330 × 10^-3 mol/mL. The amount of KOH in 1.69×10−2 mol is:

Amount of solute = 1.69×10−2 mol

Putting the values in the formula, we get:

Volume = (1.69×10−2 mol) / (0.330 × 10^-3 mol/mL)

Volume = 51.2 mL

Therefore, 51.2 mL of the 0.330 mm KOH solution contains 1.69×10−2 mol of KOH.

To learn more about Molarity click here

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