Determine the mass of ammonium chloride, NH4CI, required to prepare 0.250 L of a 0.35 M solution of ammonium chloride.

Chemistry

1 answer:

kkurt [141]3 years ago

7 0

A 0.250 L solution of NH4Cl contains (0.250 L)(0.35 mol/L) = 0.0875 moles of dissolved NH4Cl. Converting this quantity to mass, we get (0.0875 mol NH4Cl)(53.491 g/mol) = 4.7 g NH4Cl (two significant figures).

So, approximately 4.7 g of NH4Cl dissolved in a 0.250 L solution will have a concentration of 0.35 L.

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