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?

1 Answer
Nov 30, 2015

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

Explanation:

Convert mM to M.

0.45cancel"mM O"_2""xx(1"M O"_2"")/(1000cancel"mM O"_2"")="0.00045 M O"_2"

"0.00045 M O"_2"="0.00045 mol/L O"_2"

Convert 300 mL to liters.

300.0cancel"mL solution"xx(1"L")/(1000cancel"mL")="0.3000 L solution"

Determine moles of "O"_2" dissolved in "0.3000 L" of solution.

0.3000cancel"L solution"xx(0.00045"mol O"_2)/(1cancel"L solution")="0.000135 mol O"_2"

"STP"="273.15 K" and "100 kPa"

Use the Ideal Gas Law

PV=nRT, where P="pressure", V="volume", n="moles", R="gas constant", and T="Kelvin temperature".

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

Unknown
volume, V

Solution
Rearrange the equation to isolate V and solve.

V=(nRT)/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.

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