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Question

Calculate the concentration of the solution that 40.0 mL of the 22% (w/w) NH3 (17,0 g/mol) solution with a density of 0.870 g/mL is diluted to 300 mL.

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Answer 1

40.0 ml of solution x 0.870 g/ml = 34.8 g of solution.

Since 22% of this is NH₃, we have 0.22 x 34.8 g = 7.656 g NH₃

Moles of NH₃ present = 7.656 g NH₃ x 1 mol NH₃/17.0 g = 0.450 moles NH₃

Volume = 300 ml = 0.300 L

Concentration (M) of NH₃ = 0.450 moles/0.300 L = 1.50 M