The molality and mole fraction of solute in a 21.6 percent by mass aqueous solution of formic acid are 5.98mol and 0.097 respectively.
21.6 %m/m = 21.6 g HCOOH dissolved in (100 g - 21.6 g=) 78.4 g H2O
Molar mass of HCOOH = 46.03 g/mol
Mol in 21.6 g = 21.6 g / 46.03 g/mol = 0.4692 mol
78mol HCOOH dissolved in 1000g water= (1000g/78.4g)0.4692 mol
Molality= 5.98mol in 1.0 kg water.
Molar mass H2O = 18 g/mol
Mol H2O in 1000 g = 55.55 mol
Total moles in solution = 5.98mol+55.55mol= 61.53mol
Mol fraction HCOOH = 5.98 mol / 61.53 mol = 0.097
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