solution of Formic acid, is 75% HCOOH by mass & has adensity of 1.20 g/ml. Calcu

Question

solution of Formic acid, is 75% HCOOH by mass & has adensity of 1.20 g/ml. Calculate the molarity of this solution.(HINT: Assume 1-L solution). This is what i got : mol HCOOH= 1.00 L soln * 1000 mL soln/ 1L soln * 1.20 gsoln/1mL soln * 75gHCOOH/ 100g soln * 1 mol HCOOH/46.02538 = 19.5 19.5 mol HCOOH/1.00 L soln = 19.5 M HCOOH is that right ? solution of Formic acid, is 75% HCOOH by mass & has adensity of 1.20 g/ml. Calculate the molarity of this solution.(HINT: Assume 1-L solution). This is what i got : mol HCOOH= 1.00 L soln * 1000 mL soln/ 1L soln * 1.20 gsoln/1mL soln * 75gHCOOH/ 100g soln * 1 mol HCOOH/46.02538 = 19.5 19.5 mol HCOOH/1.00 L soln = 19.5 M HCOOH is that right ?

Explanation / Answer

That's what I came up with. 1.2g/ML = 1200g/L 1200g/L x 75%= 900g/L HCOOH 900g/L * 1/46.03g/mole= 19.55 mol/L so yes you got it right ;)

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