Respuesta :

mergl
1.50L of 0.500m KCl
1.50(0.500/1 L)=.75 mol KCl
.75 mol KCl(74.55 g/1 mol)=55.1925 g KCl

55.875 grams of KCl are necessary to prepare 1.50 liters of a 0.500 M solution of KCl.

Define molarity of a solution.

Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.

Molarity

Molality = [tex]\frac{Moles \;solute}{\;Volume \;of \;solution \;in \;litre}[/tex]

First, let's calculate the total number of moles in this solution:

Moles = [tex]\frac{mass}{molar \;mass}[/tex]

(molar mass of KCl is 74.5 gram/mol)

First, let's calculate the total number of moles in this solution:

Molality = [tex]\frac{Moles \;solute}{\;Volume \;of \;solution \;in \;litre}[/tex]

0.500 = [tex]\frac{Moles \;solute}{1.50}[/tex]

Moles = 0.75

Now find mass

Moles  = [tex]\frac{mass}{molar \;mass}[/tex]

0.75 = [tex]\frac{mass}{74.5}[/tex]

55.875 = mass

Hence, 55.875 grams of KCl are necessary to prepare 1.50 liters of a 0.500 M solution of KCl.

Learn more about molarity here:

https://brainly.com/question/3563925

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