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What will be the molarity of formaldehyde in water solution at \( 298 K \) if formaldehyde occurs at \( 1.5 atm \) above the solution. \( K _{ H } \) of formaldehyde at \( 298 K \) is \( 1.83 \times 10^{2} atm \). |
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Answer» According to henry's law- "The partial pressure applied by any gas on a liquid surface is directly proportional to its mole fraction present in a liquid solvent. P = Kn.X KH = henry's constant X = mole fraction of solute. ∴ 1.5 atm = 1.83 x 102 atm \(\times\) x x = \(\frac{1.5}{1.83\times10^2}\) = 0.0082 mole fraction of HCHO = 0.0082 = \(\frac{n_{CHO}}{n_{H_2o+n_{HCHO}}}\) ∵ nH2O >> nHCHO we can neglect nHCHO in denominator ∴ 0.0082 = \(\frac{n_{HCHO}}{n_{H_2O}}\) 0.0082 = \(\frac{n_{HCHO}}{\frac{W}{18}}\) or = \(\frac{n_{HCHO}}w\times1000\) = 0.0082 x 18 ⇒ molality = 0.0082 x 18 As we know for very dilute solution the molality and molarity are approximately equal. ∴ molarity of HCHO = 0.15 M |
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