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if 45.3 ml of a 0.108 m hcl solution is needed to neutralize a solution of koh, how many grams of koh must be present in the solution?

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tutorconsortium272

If 45.3 ml of a 0.108 M HCL solution is needed to neutralize a solution of KOH, then grams of KOH that  must be present in the solution is 0.274gm.

What is molarity?

The amount of a substance present in a certain volume of solution is termed as molarity . It can also be defined as the moles of a solute per liter of solution.

HCL +KOH --->  KCL+ H2O

Molarity= mol solute / L solution

Given volume of HCL= 45.3ml= 0.0453L

concentration = 0.108M

As we know that, No. of moles = concentration * Volume

=0.0453* 0.108

Number of moles=0.0048

Mass = no. of moles*molar mass

Molar mass of KOH = 56.1 g/mol

= 0.00489* 56.1

Grams of KOH= 0.274 gm.

To know more about molarity, refer

https://brainly.com/question/14469428

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