# Question #8cb12

Feb 28, 2016

${H}^{c p} = 1.8 \cdot {10}^{- 6} {\text{mol L"^(-1)"mmHg}}^{- 1}$

#### Explanation:

This is a pretty straightforward Henry's Law practice problem in which you have to use the molar solubility of the gas and the partial pressure of the gas to find the value of Henry's constant, ${H}^{c p}$.

As you know, Henry's Law states that the solubility of a gas in a liquid is proportional to that gas' partial pressure above the liquid.

$\textcolor{b l u e}{c = {H}^{c p} \cdot P} \text{ }$, where

$c$ - the concentration of the dissolved gas
${H}^{c p}$ - Henry's constant
$P$ - the partial pressure of the gas

${H}^{c p} = \frac{c}{P}$

Plug in your values to get

${H}^{c p} = {\text{0.00140 mol L"^(-1)/("760 mmHg") = "0.000001842 mol L"^(-1)"mmHg}}^{- 1}$

You need to round this off to two sig figs, since that's how many sig figs you have for the partial pressure of the gas.

Expressed in scientific notation and rounded to two sig figs, the answer will be

${H}^{c p} = \textcolor{g r e e n}{1.8 \cdot {10}^{- 6} {\text{mol L"^(-1)"mmHg}}^{- 1}}$