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Chapter 11: Problem 57

A 7.24-g sample of gas is contained in a 4.00-L flask. Its pressure is \(765.0 \mathrm{~mm} \mathrm{Hg}\), and its temperature is \(25.0^{\circ} \mathrm{C}\). What is the molar mass of this gas?

Short Answer

Expert verified
The molar mass of the gas is calculated using the Ideal Gas Law and the given conditions: given mass (7.24 g), pressure (765.0 mmHg), volume (4.00 L), and temperature (25.0°C). After converting pressure to atm (1.0066 atm) and temperature to Kelvin (298.15 K), the number of moles (n) of gas can be found (0.1642 mol). Finally, the molar mass is determined by dividing the mass of the gas by the number of moles, resulting in a molar mass of 44.1 g/mol.

Step by step solution

01

Convert pressure to atm

Given pressure is 765.0 mmHg, we'll convert it to atm because the ideal gas constant R is given in atm. 1 atm = 760 mmHg Pressure (P) = 765.0 mmHg × (1 atm / 760 mmHg) = 1.0066 atm
02

Convert temperature to Kelvin

Temperature (T) in Celsius is given as 25.0°C. To convert it to Kelvin, we'll add 273.15. Temperature (T) = 25.0°C + 273.15 = 298.15 K
03

Use Ideal Gas Law to find number of moles (n)

Now knowing the temperature T and pressure P, along with the provided volume V (4.00 L), we can use the Ideal Gas Law to solve for the number of moles: PV = nRT R = 0.0821 L atm / K mol Rearranging the equation for n, we get n = PV / RT n = (1.0066 atm)(4.00 L) / (0.0821 L atm / K mol)(298.15 K) n = 0.1642 mol
04

Calculate molar mass

Now knowing the number of moles n and the given mass (7.24 g), we can find the molar mass of the gas: Molar mass = mass / number of moles Molar mass = 7.24 g / 0.1642 mol Molar mass = 44.1 g/mol Thus, the molar mass of this gas is 44.1 g/mol.

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Most popular questions from this chapter

In the ideal gas law, pressure must be expressed in units of _______ volume must be expressed in units of _______ temperature must be expressed in _______ and \(n\) represents the number of _______

A student thinks he remembers reading that if you double the temperature of an ideal gas, its. pressure doubles. He is given a problem where he has an ideal gas at \(25.0{ }^{\circ} \mathrm{C}\) and \(2.5 \mathrm{~atm}\). He is asked what the temperature must be raised to in order to double the pressure to \(5.0 \mathrm{~atm}\). He answers, \(^{\prime \prime} 50.0{ }^{\circ} \mathrm{C}\), of course." Why is he wrong? What lesson should he learn about using the ideal gas law? What is the temperature increase in Celsius degrees that will double the pressure?

What volume in milliliters of \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas at \(0.996 \mathrm{~atm}\) and \(25.0^{\circ} \mathrm{C}\) contains \(0.200 \mathrm{~g}\) of oxygen?

Rewrite the ideal gas law solving for \(n\). Also show how all units cancel to leave you with just units of moles.

Arrange these gases, all at STP, in order of increasing density: \(\mathrm{CO}_{2}, \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CH}_{4}, \mathrm{He}\).
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