Question

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

Answer 1

Answer: the mass of O2 present in the sample under the given conditions is 0.1495g

Step-by-step explanation:

The question requires us to calculate the mass of oxygen gas (O2) that is present under the given conditions:

Volume = 120.0 mL

Pressure = 700.0 mmHg

Temperature = 15°C

To solve this problem, we can apply the equation of ideal gases, which states:

`PV=nRT`

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:

`PV=nRT\rightarrow n=(PV)/(RT)`

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:

`\begin{gathered} n=(PV)/(RT)\text{ and n =}(m)/(M)\rightarrow(m)/(M)=(PV)/(RT) \\ \\ m=(P* V* M)/(R* T) \end{gathered}`

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:

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