There are 120.0 mL of 02 at 700. 0 mmHg and 15⁰ C. What is the number of grams present?

where P refers to the pressure of the system, V is the volume of gas, R is the constant of gases (R = 0.0821 L.atm/K.mol) and T is the temperature of the gas. Note that, according to the units used for the constant of gases, we need to use pressure in units of atm, V in units of liter, n is given in moles and T must be used in Kelvin.

We can rearrange the equation above to calculate the number of moles of the gas:

[tex]PV=nRT\rightarrow n=\frac{PV}{RT}[/tex]

And, knowing that the number of moles of a susbtance corresponds to the ratio between the mass of the sample (m) and the molar mass of the compound (M), we can write:

[tex]\begin{gathered} n=\frac{PV}{RT}\text{ and n =}\frac{m}{M}\rightarrow\frac{m}{M}=\frac{PV}{RT} \\ \\ m=\frac{P\times V\times M}{R\times T} \end{gathered}[/tex]

Before applying the equation above, we need to convert the given values of pressure, volume and temperature to the appropriate units:

- Pressure:

1 atm corresponds to 760 mmHg, thus:

760 mmHg ----------- 1 atm

700.0 mmHg ------- x

Solving for x, we have that 700.0 mmHg corresponds to 0.9211 atm.

- Volume:

1 L corresponds to 100 mL, thus:

1000 mL ----------- 1 L

120.0 mL ---------- y

Solving for y, we have that 120.0 mL corresponds to 0.1200 L.

- Temperature:

0°C corresponds to 273.15 K, thus we can add 273.15 to the given temperature value to obtain the corresponding temperature:

T(K) = 15 + 273.15 = 288.15 K

Therefore, 15°C corresponds to 288.15 K.

Now that we have all values in the correct unit (P = 0.9211 atm, V = 0.1200 L and T = 288.15 K), we can apply these values to our rearranged equation to obtain the mass of O2. The molar mass of O2 is 31.998 g/mol, thus we have:

[tex][/tex]

Therefore, the mass of O2 present in the sample under the given conditions is 0.1495g.