How many moles of each substance are needed to prepare the following solutions?

45.0 ml of 7.5 %m/v kcl (mw = 74.55 g/mol)

400.0 ml of 9.5 %m/v acetic acid (mw = 60.05 g/mol)

A) 7.5 % m / V 

7.5 g 100 mL ( solution )
mass g 45.0 mL

mass g = 45.0 x 7.5 / 100

mass g = 337.5 / 100 => 3.375 g

1 mole KCl 74.55 g
moles KCl 3.375 g

moles KCl = 3.375 x 1 / 74.55 => 0.0452 moles of KCl


b) 9.5 % ( m / V ) :

9.5 g 100 mL ( solution )
mass g 400.0 mL

mass g = 400.0 x 9.5 / 100

mass g = 3800 / 100 => 38 g

1 mole acetic acid  60.05 g
moles acetid acid   38 g

moles acetid acid = 38 x 1 / 60.05 => 0.6328 moles of  acetid acid 


hope this helps!

The answer is c fish