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How many milliliters of a 0.619 M KOH solution would be needed to provide 5.36 g of KOH to a chemical reaction?

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Final answer:

To provide 5.36 g of KOH using a 0.619 M KOH solution, one would need to calculate the moles of KOH required and then use the molarity to find the volume, which comes out to approximately 154 mL.

Step-by-step explanation:

To calculate how many milliliters of a 0.619 M KOH solution would be needed to provide 5.36 g of KOH, you first need to determine how many moles of KOH you have using its molar mass.

One mole of KOH has a molar mass of approximately 56.11 g/mol. Using the formula:

moles = mass (g) / molar mass (g/mol),

moles of KOH = 5.36 g / 56.11 g/mol = 0.0955 mol.

Now that we know the moles of KOH, we use the molarity (M) of the solution to find the volume:

Volume (L) = moles / molarity (mol/L),

Volume (L) = 0.0955 mol / 0.619 mol/L = 0.154 L,

which converts to milliliters (1 L = 1000 mL):

Volume (mL) = 0.154 L × 1000 mL/L = 154 mL.

Therefore, you would need approximately 154 mL of the 0.619 M KOH solution to provide 5.36 g of KOH for the chemical reaction.

User Anton Holovin
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