Question

Partial pressure of ethane over a saturated solution containing 6.56x10-2 gram of ethane is 1 bar. What will be the partial pressure of the gas if it contains 5x 10-2 grams of ethane.

Answer 1

Henry's law states that :

$p=k_H\ c$

where p = partial pressure of the gas above the solution.
   $k_H]$is the Henry's solubility constant for the solvent & solute at given T°C
   c = molar concentration  of the solute in the solution in  moles/Litre  at given T°C
     c = m / V ,    m = mass of the gas  and  V = volume of the solvent

So solubility of directly proportional to the partial pressure of the gas over the  solution.  Since solubility is directly proportional to the mass of the gas solute, the mass dissolved is directly proportional to the partial pressure.

 p1 = 1 bar              m1  = 6.56 * 10⁻² gm
 p2  = ?                  m2 =  5 * 10⁻²  gm

$\frac{p_2}{p_1}=\frac{m_2}{m_1}\\\\Hence,\ p_2=\frac{m_2}{m_1}*p_1=0.762\ bar$

The partial pressure is less than 1 bar which is for saturated solution.

Answer 2

Hey !!

According to Henry's Law, m = KH × p

CASE - 1

           6.56 × 10⁻²g = KH × 1 bar

(OR)          KH = 6.56 × 10⁻² g bar⁻¹

CASE - 2

           5.00 × 10⁻² g = ( 6.56 × 10⁻² g bar⁻¹) × p

(OR)  

               p = 5.00 × 10⁻² g / 6.56 × 10⁻² g bar⁻¹

FINAL RESULT = 0.762 bar

Hope it helps you !!