Question

Given the concentration of neon as $C_{Ne}=23.5mL/(L\text{of water)}$ at STP. Then calculate the Henry's constant at STP for neon dissolved in water. Given STP molar volume is $22414mL/mol$ and pressure is $1atm$.

• A. $1.048\times {10}^{-3}mol{L}^{-1}at{m}^{-1}$
• B. $3.557\times {10}^{-2}mol{L}^{-1}at{m}^{-1}$
• C. $5.55\times {10}^{-4}mol{L}^{-1}at{m}^{-1}$
• D. $0.114\times {10}^{-4}mol{L}^{-1}at{m}^{-1}$

Answer 1

The correct option is A $1.048\times {10}^{-3}mol{L}^{-1}at{m}^{-1}$

$\text{Concentration of Ne in water}=23.5mL/L$
$Numberofmoles=\frac{Volume}{Molarvolume}$

$1L$ of solution contains $23.5mL$ of neon which is equal to
$23.5\times \frac{1}{22414}=0.001048mol\text{at STP.}$
Thus, the concentration of the solution is $0.001048M=1.048\times {10}^{-3}M$
According to Henry's law, $S=K_H^{\prime }P$
Where,
$S$ is the solubility in $mol{L}^{-1}$
$K_H^{\prime }$ is the henry's law constant having units of $mol{L}^{-1}at{m}^{-1}$
$P$ is the pressure in atm.
Substituting values in the above expression.
$1.048\times {10}^{-3}=K_H^{\prime }\times 1$

$K_H^{\prime }=1.048\times {10}^{-3}mol{L}^{-1}at{m}^{-1}$