# A glass of cold water contains .45 mM O_2. How many millilitres of oxygen gas at STP are dissolved in 300.0 mL of this water?

Nov 30, 2015

There will be 3.1 mL of oxygen gas dissolved in 300 mL of solution.

#### Explanation:

Convert mM to M.

$0.45 \cancel{\text{mM O"_2""xx(1"M O"_2"")/(1000cancel"mM O"_2"")="0.00045 M O"_2}}$

$\text{0.00045 M O"_2}$$=$$\text{0.00045 mol/L O"_2}$

Convert 300 mL to liters.

$300.0 \cancel{\text{mL solution"xx(1"L")/(1000cancel"mL")="0.3000 L solution}}$

Determine moles of $\text{O"_2}$ dissolved in $\text{0.3000 L}$ of solution.

$0.3000 \cancel{\text{L solution"xx(0.00045"mol O"_2)/(1cancel"L solution")="0.000135 mol O"_2}}$

$\text{STP"="273.15 K" and "100 kPa}$

Use the Ideal Gas Law

$P V = n R T$, where $P = \text{pressure}$, $V = \text{volume}$, $n = \text{moles}$, $R = \text{gas constant}$, and $T = \text{Kelvin temperature}$.

Given/Known
$P = \text{100 kPa}$
$n = \text{0.000135 mol O"_2}$
$R = \text{8.3144598 L kPa K"^(-1) "mol"^(-1)}$
$T = \text{273.15 K}$

Unknown
volume, $V$

Solution
Rearrange the equation to isolate $V$ and solve.

$V = \frac{n R T}{P}$

V=(0.000135cancel"mol O"_2xx8.3144598"L" cancel("kPa") cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K")/(100cancel"kPa")="0.00307 L O"_2"

Convert liters to milliliters.

$0.00307 \cancel{\text{L O"_2xx(1000"mL")/(1cancel"L")="3.1 mL O"_2}}$ (rounded to two significant figures)