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How many grams of KOH are there in 250 milliliters of 0.05m aqueous solution of this substance

User Cardern
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Step-by-step explanation:

Also, we know that molarity is the number of moles present in a liter of solution.

Molarity =
\frac{\text{no. of moles}}{volume}

0.05 m =
\frac{\text{no. of moles}}{0.25 L} (as 1 ml = 0.001 L)

no. of moles = 0.0125 mol

It is known that number of moles equal mass divided by molar mass. And, molar mass of KOH is 56.10 g/mol.

Hence, No. of moles =
\frac{mass}{\text{molar mass}}

0.0125 mol =
(mass)/(56.10 g/mol)

mass = 0.701 g

Thus, we can conclude that there are 0.701 g of KOH are there in 250 milliliters of 0.05m aqueous solution of this substance.

User Nick Zoum
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