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Calculate the volume and mass required to make 0.1 M solutions of Ammonium Chloride (53.49 g/mol)...

Calculate the volume and mass required to make 0.1 M solutions of Ammonium Chloride (53.49 g/mol) and Ammonia (1.0 M

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Answer #1

basis ---> 1 liter of solution

so...

if molarity of NH4Cl --> 0.1M

this means 0.1 mol of NH4Cl are present

so

mass = mol*MW = 0.1*53.49 = 5.349 g required

(per liter!!)

for ammonia (NH3)

0.1 M of NH3

0.1 mol of NH3

MW NH3 = 17 g/mol

mass = mol*MW = 0.1*17 = 1.7 g of NH3 required

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