Question

The vapor pressure of \(\mathrm{I}_{2}(s)\) at \(30^{\circ} \mathrm{C}\) is \(0.466 \mathrm{~mm} \mathrm{Hg}\). (a) How many milligrams of iodine will sublime into an evacuated 750.0 -mL flask? (b) If \(3.00 \mathrm{mg}\) of \(\mathrm{I}_{2}\) are used, what will the final pressure in the flask be? (c) If \(7.85 \mathrm{mg}\) of \(\mathrm{I}_{2}\) are used, what will the final pressure in the flask be?

Short answer

Question: Calculate the mass of iodine that will sublime into an evacuated flask and predict the final pressure in the flask for the given masses of iodine: 3.00 mg and 7.85 mg. Answer: The mass of iodine that will sublime into the evacuated flask is 4.74 mg. The final pressure in the flask will be 0.298 mmHg for 3.00 mg of iodine and 0.774 mmHg for 7.85 mg of iodine.

Step-by-step solution

Part (a): Calculate mass of Iodine that will sublime

First, we need to find the number of moles of Iodine that will sublime into the empty flask. The given vapor pressure of solid iodine is 0.466 mmHg. For that, we will use the ideal gas law equation: \(PV = nRT\) Here, P = Vapor pressure V = Volume of the flask n = number of moles R = Ideal gas constant T = Temperature (in Kelvin) We need to find the number of moles (n) that will sublime into the flask. Given values: Pressure P = 0.466 mmHg To convert to atm, we use the conversion factor, 1 atm = 760 mmHg \(P = 0.466/760 = 0.000613\ \mathrm{atm}\) Volume V = 750.0 mL To convert to liters, we use the conversion factor, 1 L = 1000 mL \(V = 750.0/1000 = 0.75\ \mathrm{L}\) Temperature T = \(30^\circ\mathrm{C}\) To convert to Kelvin, add 273.15 \(T = 30 + 273.15 = 303.15\ \mathrm{K}\) and the Ideal gas constant R = 0.0821 atm.L/(mol.K) Now, substitute the given values into the equation (PV=nRT) and solve for 'n': \(n = \frac{PV}{RT} = \frac{(0.000613\ \mathrm{atm})(0.75\ \mathrm{L})}{(0.0821\ \mathrm{atm\cdot L/(mol\cdot K)})(303.15\ \mathrm{K})} = 1.87 \times 10^{-5}\ \mathrm{mol}\) Now, we will calculate the mass of iodine: Molar mass of Iodine (\(\mathrm{I}_2\)) = 253.8 g/mol Mass = number of moles × molar mass \(Mass = 1.87 \times 10^{-5}\ \mathrm{mol} \times 253.8\ \mathrm{g/mol}\) To convert to milligrams, use the conversion factor, 1 g = 1000 mg Substituting and evaluating yields: \(Mass = (1.87 \times 10^{-5}\ \mathrm{mol})(253.8\ \mathrm{g/mol})(1000\ \mathrm{mg/g}) = 4.74\ \mathrm{mg}\) Thus, 4.74 mg of iodine will sublime into the evacuated flask.

Part (b): Final pressure in the flask when 3.00 mg of Iodine is used

First, we need to find the number of moles of iodine (n) in 3.00 mg. For that, we will use the molar mass of iodine and the given mass: Mass = 3.00 mg To convert to grams, use the conversion factor, 1 g = 1000 mg Mass = 3.00 / 1000 = 0.003 g Molar mass of Iodine (\(\mathrm{I}_2\)) = 253.8 g/mol Now, calculate the number of moles: \(n = \frac{Mass}{Molar\ Mass} = \frac{0.003}{253.8} = 1.18 \times 10^{-5}\ \mathrm{mol}\) Next, we will use the number of moles and the ideal gas law equation to calculate the final pressure in the flask. Given: Volume V = 0.75 L Temperature T = 303.15 K Ideal gas constant R = 0.0821 atm.L/(mol.K) Now, substitute the given values into the equation (PV = nRT) and solve for 'P': \(P = \frac{nRT}{V} = \frac{(1.18 \times 10^{-5}\ \mathrm{mol})(0.0821\ \mathrm{atm\cdot L/(mol\cdot K)})(303.15\ \mathrm{K})}{0.75\ \mathrm{L}} = 0.000392\ \mathrm{atm}\) To convert the pressure to mmHg, we use the conversion factor, 1 atm = 760 mmHg: \(P = 0.000392\ \mathrm{atm} \times 760 = 0.298\ \mathrm{mmHg}\) Thus, the final pressure in the flask will be 0.298 mmHg when 3.00 mg of Iodine is used.

Part (c): Final pressure in the flask when 7.85 mg of Iodine is used

The process in part (c) is similar to part (b). First, we find the number of moles (n) of Iodine in 7.85 mg and then use the ideal gas law equation to find the final pressure. Mass = 7.85 mg To convert to grams, use the conversion factor, 1 g = 1000 mg Mass = 7.85 / 1000 = 0.00785 g Molar mass of Iodine (\(\mathrm{I}_2\)) = 253.8 g/mol Now, calculate the number of moles: \(n = \frac{Mass}{Molar\ Mass} = \frac{0.00785}{253.8} = 3.09 \times 10^{-5}\ \mathrm{mol}\) Given: Volume V = 0.75 L Temperature T = 303.15 K Ideal gas constant R = 0.0821 atm.L/(mol.K) Now, substitute the given values into the equation (PV = nRT) and solve for 'P': \(P = \frac{nRT}{V} = \frac{(3.09 \times 10^{-5}\ \mathrm{mol})(0.0821\ \mathrm{atm\cdot L/(mol\cdot K)})(303.15\ \mathrm{K})}{0.75\ \mathrm{L}} = 0.00102\ \mathrm{atm}\) To convert the pressure to mmHg, we use the conversion factor, 1 atm = 760 mmHg: \(P = 0.00102\ \mathrm{atm} \times 760 = 0.774\ \mathrm{mmHg}\) Thus, the final pressure in the flask will be 0.774 mmHg when 7.85 mg of Iodine is used.