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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)

Respuesta :

Willchemistry
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
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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 
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hope this helps!


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