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Chapter 6: Problem 70

A 1.072 g sample of \(\mathrm{He}(\mathrm{g})\) is found to occupy a volume of 8.446 L when collected over hexane at \(25.0^{\circ} \mathrm{C}\) and \(738.6 \mathrm{mmHg}\) barometric pressure. Use these data to determine the vapor pressure of hexane at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The vapor pressure of hexane at 25.0°C is \(0.1\) atm. Remember, results may need onward conversion if other units are required.

Step by step solution

01

Convert units and conditions

First, convert all the units to their standard counterparts. Mass of helium (He) = 1.072g. The molar mass of helium is 4.00 g/mol, so the moles of He is given by 1.072 g / 4.00 g/mol = 0.268 mol. The volume of He is given as 8.446 L. Temperature is given in Celsius, so it must be converted into Kelvin by adding 273.15. So, T = 25.0 °C + 273.15 = 298.15 K. The barometric pressure is given in mmHg; to convert it into atmospheres, divide by 760. So, P = 738.6 mmHg / 760 = 0.972 atm.
02

Apply the ideal gas law

We assume the total pressure is the sum of the pressure of He and the vapor pressure of hexane by Dalton's law of partial pressures. We first need to find the pressure contributed by helium using the ideal gas law. Plugging in the values into PV = nRT, we get (P_He)(8.446 L) = (0.268 mol)(0.0821 L·atm/(K·mol))(298.15 K). Solving for P_He, we discover it equals to 0.872 atm.
03

Calculate the vapor pressure of hexane

According to Dalton's law, the total pressure is the sum of the pressure of He and the vapor pressure of hexane (P_hexane). So, P_hexane + P_He = P_total. Rearrange to find P_hexane = P_total - P_He. Plugging in the values we have, P_hexane = 0.972 atm - 0.872 atm = 0.1 atm. Note that the pressure units are in atm, so if required in another unit, appropriate conversion is necessary.

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Most popular questions from this chapter

A 1.65 g sample of \(\mathrm{Al}\) reacts with excess \(\mathrm{HCl}\), and the liberated \(\mathrm{H}_{2}\) is collected over water at \(25^{\circ} \mathrm{C}\) at a barometric pressure of \(744 \mathrm{mmHg} .\) What volume of gaseous mixture, in liters, is collected? $$2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{g})$$

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