The partial pressure of nitrogen in air is 0.76 atm. and its Henry’s law constant is 7.6 × 104 atm at 300 K. What is the mole fraction of N2 in the solution obtained when air is bubbled through water at 300 K?

Question

The partial pressure of nitrogen in air is 0.76 atm. and its Henry’s law constant is 7.6 × 104 atm at 300 K. What is the mole fraction of N2 in the solution obtained when air is bubbled through water at 300 K? (A) 1 × 10-4 (B) 2 × 10-4 (C) 1 × 10-5 (D) 2 × 10-5

Tardigrade Admin · Accepted Answer

PN2=0.76 atm K H =7.6 × 104 x =? PN2= K ⋅ x 0.76=7.6 × 104 × x =0.76 /(7.6 × 104)=1 × 10-5

Q. The partial pressure of nitrogen in air is $0.76 a t m$ . and its Henry’s law constant is $7.6 \times 1 0^{4}$ atm at 300 K. What is the mole fraction of $N_{2}$ in the solution obtained when air is bubbled through water at 300 K?

$1 \times 1 0^{- 4}$

$2 \times 1 0^{- 4}$

$1 \times 1 0^{- 5}$

$2 \times 1 0^{- 5}$

$1 \times 1 0^{- 6}$

$P_{N_{2}} = 0.76 a t m$
$K_{H} = 7.6 \times 1 0^{4}$
$x = ?$
$P_{N_{2}} = K \cdot x$
$0.76 = 7.6 \times 1 0^{4} \times$
$x = 0.76/ (7.6 \times 1 0^{4}) = 1 \times 1 0^{- 5}$

Q. The partial pressure of nitrogen in air is 0.76atm. and its Henry’s law constant is 7.6×104 atm at 300 K. What is the mole fraction of N2​ in the solution obtained when air is bubbled through water at 300 K?

1×10−4

2×10−4

1×10−5

2×10−5

1×10−6

PN2​​=0.76atm KH​=7.6×104 x=? PN2​​=K⋅x 0.76=7.6×104× x=0.76/(7.6×104)=1×10−5