Question

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constant for Ar under these conditions is 1.5 × 10−3 mol/L·atm. Enter your answer in scientific notation.

Answer 1

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

where S: is the solubility or concentration of Ar in water, $k_{H}$: is Henry's law constant and p: is the pressure of the Ar

Since the argon is 0.93%, we need to multiply the equation (1) by this percent:

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

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