Question

How many grams of solid potassium chlorate (KClO 3, 122.55 g/mol) are needed to make 150 mL of 0.50 M solution

Answer 1

`m_(solute)=9.19gKClO_3`

Step-by-step explanation:

Hello,

In this case, since the molarity is defined as:

`M=(mol_(solute))/(V_(solution))`

Whereas the volume of the solution is in liters, we first compute the moles of solute which in this case is the potassium chlorate:

`mol_(solute)=150mL*(1L)/(1000mL)*0.50molKClO_3/L \\\\mol_(solute)=0.075molKClO_3`

Then, by using its molar mass of 122.55 g/mol, we compute the required mass as follows:

`m_(solute)=0.075molKClO_3*(122.55gKClO_3)/(1molKClO_3) \\\\m_(solute)=9.19gKClO_3`

Regards.