Sulfamic acid, HSO3NH2 (molar mass = 97.1 g/mol), is a strong monoprotic acid that can be used to standardize a strong base: A 0.179-g sample of HSO3NH2 required 19.4 mL of an aqueous solution of KOH for a complete reaction. What is the molarity of the KOH solution?

The molarity of the KOH solution is 0.093 M

We'll begin by calculating the number of mole of HSO₃NH₂.

Mole = mass / molar mass

Mole of HSO₃NH₂ = 0.179 / 97.1

Mole of HSO₃NH₂ = 0.0018 mole

Next, we shall determine the number of mole of KOH required to react with 0.0018 mole of HSO₃NH₂.

HSO₃NH₂ + KOH —> KSO₃NH₂ + H₂O

From the balanced equation,

1 mole of HSO₃NH₂ reacted with 1 mole of KOH.

Therefore,

0.0018 mole of HSO₃NH₂ will also react with 0.0018 mole of KOH.

Finally, we shall determine the molarity of the KOH solution.

Molarity = mole / Volume

Molarity of KOH = 0.0018 / 0.0194

Molarity of KOH = 0.093 M

Thus, the molarity of the KOH solution is 0.093 M.

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The molarity of KOH solution is 0.093 M

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.

where,

n = moles of solute

= volume of solution in ml

moles of =

Now the balanced chemical equation will be :

As 1 mole of requires 1 mole of KOH

Thus 0.0018 moles of requires = moles of KOH

Now put all the given values in the formula of molality, we get

Therefore, the molarity of KOH solution is 0.093 M

it's filtration, but this is an assumption so my apologies if i got this

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explanation: