Question

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 \( 1.83 \times 10^{2} atm \).

Answer 1

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 = K_n.X

K_H = 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 >> n_HCHO

we can neglect n_HCHO 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