Question

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

Answer 1

Vs = 1.0 mL = 0.001 L
c((NH4)2CO3) = 0.02 M
n((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