Question

the partial pressure of ethane over a solution containing 6.56*ten power(-3)g of ethane is 1 bar. if the solution contains 5.00*ten power(-2)g of ethane,then what shall be the partial pressure f the gas?

Answer 1

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 !!

Answer 2

Answer:

Explanation:According to Henry’s law,

                      We know that,m=KH*p

                     5.00*10 to the power -2=(6.56*10 to the power -3)*p

P=5.00*10to the power -2/6.56*10 to the power -3

P=0.762bar