The Henry law constant for dissolution of a gas in aqueous medium is 3× 102atm . At what partial pressure of the gas the molality of aqueous solution will be (5/9)m :

Question

The Henry law constant for dissolution of a gas in aqueous medium is 3× 102atm . At what partial pressure of the gas the molality of aqueous solution will be (5/9)m : (A) 1.5atm (B) 3atm (C) 6atm (D) 9atm

Tardigrade Admin · Accepted Answer

P=KH((ng/ng + nl)) but ng < < n1P=KH((ng/nl)) =KH((m/1000) × M textsolvent ) =3× 102× (5/9)× (1/1000)× 18=3atm

Q. The Henry law constant for dissolution of a gas in aqueous medium is $3 \times 1 0^{2} a t m$ . At what partial pressure of the gas the molality of aqueous solution will be $\frac{5}{9} m$ :

$1.5 a t m$

$3 a t m$

$6 a t m$

$9 a t m$

$P = K_{H} (\frac{n _{g}}{n _{g} + n _{l}})$ but $n_{g} << n_{1} P = K_{H} (\frac{n _{g}}{n _{l}})$
$= K_{H} (\frac{m}{1000} \times M_{solvent})$
$= 3 \times 1 0^{2} \times \frac{5}{9} \times \frac{1}{1000} \times 18 = 3 a t m$

Q. The Henry law constant for dissolution of a gas in aqueous medium is 3×102atm . At what partial pressure of the gas the molality of aqueous solution will be 95​m :

1.5atm

3atm

6atm

9atm