Question

At $300K,40mL$ of $O_3\left(g\right)$ dissolves in $100g$ of water at $1.0atm$. What mass of ozone dissolved in $400g$ of water pressure of $4.0atm$ at $300K$ ?

• A. $0.1g$
• B. $1.24g$
• C. $0.48g$
• D. $4.8g$

Answer 1

The correct option is D $4.8g$

According to Henry's law, concentration of dissolved gas is directly proportional to the pressure.
$C=KP$
$\frac{C^{\prime }}{C}=\frac{P^{\prime }}{P}$

In this example, we will express concentration in terms of volume of dissolved ozone divided by mass of water

Let x mL of ozone dissolve in 400 g of water at 4.0 atm pressure.

$\frac{\frac{xmL}{400g}}{\frac{40mL}{100g}}=\frac{4.0atm}{1.0atm}$
$\frac{x}{160}=4.0$
$x=640mL$

Convert unit from mL to L

$x=640mL\times \frac{1L}{1000mL}$
$x=0.640L$

Calculate number of moles of ozone.

$n=\frac{PV}{RT}$
$n=\frac{4.0atm\times 0.640L}{0.08206Latm/mol/K\times 300K}$
$n=0.10mol$

The molar mass of ozone is 48 g/mol

Mass of ozone $=48g/mol\times 0.10mol=4.8g$