The solubility of N2 in blood at 37 °C and at a partial pressure of 0.80 atm is 5.6×10−4mol L−1. The Henry’s law constant for this system is

The solubility of $N_{2}$ in blood at 37 °C and at a partial pressure of $0.80 a t m$ is $5.6 \times 10^{- 4} mol L^{- 1}$ . The Henry's law constant for this system is

A
$1.43 \times 10^{3} atm L {mol}^{- 1}$
B
$\begin{aligned} 1.62 \times 10^{3} atm L {mol}^{- 1} \end{aligned}$
C
$5.6 \times 10^{- 4} atm L {mol}^{- 1}$
D
$2.86 \times 10^{2} atm L {mol}^{- 1}$

Solution:

Henry's law in terms of concentration of dissolved gas is $p = k_{H} c$ . Hence,

$k_{H} = \frac{p}{c} = \frac{0.80 atm}{5.6 \times 10^{- 4} mol L^{- 1}} = 1.43 \times 10^{3} atm L {mol}^{- 1}$

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