Question

The vapour pressure of water is 12.3 k pa at 300 k. Calculated the vapour pressure of molal solution in it.

Options

• 12.08 k pa

• 11.76 k pa

• 24.42 k pa

• 82.81 k pa

Answer 1

12.08 k pa

Explanation:

1 molal solution implies one mole of the solute dissolved in 1000 g (1 kg) of solvent water no. of moles of solute.

No. of moles of water of Solute (nA) = `"Mass water"/"Molar mass" = (1000  g)/(18  g  mol^-1)` = 55.55 mol

No. of moles of water (NA) = `"Mass water"/"Molar mass"` = 5.55

No. of moles of solute (MB) = 1

Mole fraction of selute (xB) = `(mB)/(mB + nA) = (1  mol)/(1  mol + 55.5  mol) = 1/56.55` = 0.0177

Vapaur pressure of so in (PA) = PÅxA = PÅ (1 – xB) = 12.3 k pa × (1 – 0.0177)

= 12.3 Ka × 0.9823

= 12.08 k pa.