Question

An open vessel containing (a) water, (b) benzene, (c) mercury stands in a laboratory measuring 6.0 m × 5.3 mx3.2 m at 25°C. What mass of each substance will be found in the air if there is no ventilation? (The vapor pressures are (a) 3.2kPa, (b) 14kPa,(c) 0.23 Pa.)

Answer 1

Solution: Find the mass in each substance

Follow the formula $PV = nRT$

but $n = \frac{m}{M}$

Then $PV = \frac{m}{M}RT \Rightarrow m = \frac{PVM}{RT}$

 

Parameter	Value
V	101.76 m3
R	8.314 J/(mol·K)
T	298 K

 

a. mass of water (H₂O)

since $P = 3.2 kPa = 3.2 \times 10^3 Pa$, $M = 18 g/mol$

Then $m_{\text{water}} = \frac{3.2 \times 10^3 \times 101.76 \times 18}{8.31 \times 298} = 2366.9 g$

 

Therefore, the mass of water is	2366.9 g

 

b. mass of benzene (C₆H₆)

since $P = 14 kPa = 14 \times 10^3 Pa$, $M = 78 g/mol$

Then $m_{\text{benzene}} = \frac{14 \times 10^3 \times 101.76 \times 78}{8.31 \times 298} = 44872.72 g

 

Therefore, the mass of benzene is	44872.72 g

 

c. mass of mercury (Hg)

since $P = 0.23 Pa$, $M = 200 g/mol$

Then $m_{\text{mercury}} = \frac{0.23 \times 101.76 \times 200}{8.31 \times 298} = 1.89 g

 

Therefore, the mass of mercury is	1.89 g

 

a. Therefore, the mass of water is 2366.9 g

b. Therefore, the mass of benzene is 44872.72 g

c. Therefore, the mass of mercury is 1.89g