3 years ago

How many moles of solute particles are present in 1 ml of aqueous 0.020 m (nh4)2co3?

1 answer:

6 0

Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = <span>0.02 M
n(</span>(NH4)2CO3) = ?

For the purpose, here we will use the next equation:

c=n/V ⇒ n=cxV

n((NH4)2CO3) = 0.02M x 0.001L

n((NH4)2CO3) = 2x10⁻⁵ mole of (NH4)2CO3 is presented in the solution

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