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Can you please help me with this chemistry problem?
Ascorbis acid (vitamin C, C6H8O6) is a water-soluable vitamin. A solution containing 80.5 g of absorbic acid dissolved in 210.0 g of water has a density of 1.22 g/L at 55 degrees celcius
Can you please show me how to calculate the mass percentage, the mole fraction, the molality, and the molarity of absorbic acid in the solution?
1 個解答
- Roger the MoleLv 75 年前最愛解答
mass percent:
(80.5 g ascorbic acid) / (80.5 g + 210.0 g) = 0.277 = 27.7% ascorbic acid by mass
mole fraction:
(80.5 g C6H8O6) / (176.1241 g C6H8O6/mol) = 0.45706 mol C6H8O6
(210.0 g H2O) / (18.01532 g H2O/mol) = 11.6567 mol H2O
(0.45706 mol C6H8O6) / (0.45706 mol + 11.6567 mol) = 0.0377 [the mole fraction of ascorbic acid]
molality:
(0.45706 mol C6H8O6) / (0.2100 kg H2O) = 2.18 mol/kg = 2.18 m ascorbic acid
molarity:
"1.22 g/L" is absurd as the density of a liquid (most gases are more dense than that), so supposing you meant "1.22 g/mL":
(80.5 g + 210.0 g) / (1.22 g/mL) = 238.11 mL = 0.23811 L of solution
(0.45706 mol C6H8O6) / (0.23811 L) = 1.92 mol/L = 1.92 M ascorbic acid
