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

Answer:

$m_{solute}=9.19gKClO_3$

Explanation:

Hello,

In this case, since the molarity is defined as:

$M=\frac{mol_{solute}}{V_{solution}}$

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

$mol_{solute}=150mL*\frac{1L}{1000mL}*0.50molKClO_3/L \\\\mol_{solute}=0.075molKClO_3$

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

$m_{solute}=0.075molKClO_3*\frac{122.55gKClO_3}{1molKClO_3} \\\\m_{solute}=9.19gKClO_3$

Regards.