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Chapter 12: Problem 29

Equilibrium is established between \(\mathrm{Br}_{2}(\mathrm{l})\) and \(\mathrm{Br}_{2}(\mathrm{g})\) at \(25.0^{\circ} \mathrm{C} .\) A \(250.0 \mathrm{mL}\) sample of the vapor weighs 0.486 g. What is the vapor pressure of bromine at \(25.0^{\circ} \mathrm{C},\) in millimeters of mercury?

Short Answer

Expert verified
The vapor pressure of Bromine at \(25.0^{\circ} C\) is approximately 71.44 mmHg.

Step by step solution

01

Find the number of moles (n)

We can find the number of moles (n) of Bromine from the given weight (0.486 g) and its molar mass. We know that Br2 has a molar mass of approximately 159.808 g/mol. Therefore, the number of moles of Br2 is \(0.486 g / 159.808 g/mol = 0.00304 mol\)
02

Convert units for volume and temperature

We need to express all quantities in the units used in the gas law equation. Therefore, convert the volume from mL to L (where 1L = 1000 mL), so we have \(V = 250 mL = 0.250 L\). Also, convert the temperature from Celsius to Kelvin (K = C + 273.15), so \(T = 25.0C = 298.15 K\)
03

Calculate the pressure

We now substitute all the derived quantities into the ideal gas law equation \(PV = nRT\). To find the pressure (P), we rearrange this as \(P = nRT/V\). Using the gas constant R = 0.0821 L.atm/mol.K, we find \(P = 0.00304 mol * 0.0821 L.atm/mol.K * 298.15 K / 0.250 L = 0.094 atm\)
04

Convert pressure to millimeters of mercury

We know that 1 atm = 760 mmHg, hence we can convert the pressure to millimeters of mercury, \(P = 0.094 atm * 760 mmHg/atm = 71.44 mmHg\)

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