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Chapter 10: Problem 51

\begin{equation}\begin{array}{l}{\text { (a) Calculate the density of } \mathrm{NO}_{2} \text { gas at } 0.970 \text { atm and } 35^{\circ} \mathrm{C} \text { . }} \\ {\text { (b) Calculate the molar mass of a gas if } 2.50 \mathrm{g} \text { occupies } 0.875} \\ {\text { L at } 685 \text { torr and } 35^{\circ} \mathrm{C} \text { . }}\end{array}\end{equation}

Short Answer

Expert verified
The density of NO2 gas at 0.970 atm and 35°C is approximately 1.85 g/L. The molar mass of the unknown gas is approximately 77.64 g/mol.

Step by step solution

01

(Part a - Step 1: Convert temperature to Kelvin)

Replace Celsius with Kelvin for the temperature. To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature. Temperature in Kelvin: \(T = 35^{\circ}\text{C} + 273.15 = 308.15 \text{K}\)
02

(Part a - Step 2: Use the Ideal Gas Law formula)

Use the Ideal Gas Law formula to find the density. The Ideal Gas Law can be expressed as: \(PV = nRT\) Where: P = Pressure in atm V = Volume of the gas in liters n = Number of moles of the gas R = Ideal Gas constant (0.0821 L atm/mol K) T = Temperature in Kelvin To find the density, we'll rearrange the formula to solve for n/V \(\frac{n}{V} = \frac{P}{RT}\)
03

(Part a - Step 3: Compute the density of NO2 gas)

Insert the values for pressure and temperature, and the Ideal Gas constant into the formula, then calculate the number of moles per volume: \(\frac{n}{V} = \frac{0.970 \text{ atm}}{(0.0821 \text{ L atm/mol K})(308.15 \text{ K})} = 0.0403 \text{ mol/L}\) Since the density of the gas is the mass of the gas divided by its volume, and the molecular mass of NO2 is 46.01 g/mol, we can calculate the density as follows: Density = \((0.0403 \text{ mol/L}) (46.01 \text{ g/mol})\) = \(1.85 \text{ g/L}\) Thus, the density of NO2 gas at 0.970 atm and 35°C is approximately 1.85 g/L.
04

(Part b - Step 1: Convert temperature and pressure to appropiate units)

Convert temperature from Celsius to Kelvin just like part (a). Temperature in Kelvin: \(T = 35^{\circ}\text{C} + 273.15 = 308.15 \text{K}\) Convert pressure from torr to atm: Pressure in atm: \(P = 685 \text{ torr} \times \frac{1 \text{ atm}}{760 \text{ torr}} = 0.901 \text{ atm}\)
05

(Part b - Step 2: Use Ideal Gas Law to find the number of moles)

Insert the values for pressure, volume, temperature and the Ideal Gas constant into the Ideal Gas Law formula, then calculate the number of moles: \(PV = nRT\) \(n = \frac{PV}{RT} = \frac{(0.901 \text{ atm})(0.875 \text{ L})}{(0.0821 \text{ L atm/mol K})(308.15 \text{ K})} = 0.0322 \text{ mol}\)
06

(Part b - Step 3: Calculate the molar mass of the gas)

Now that we have the number of moles and the mass of the gas (2.50 g), we can calculate the molar mass by dividing the mass by the number of moles: Molar mass = \(\frac{2.50 \text{ g}}{0.0322 \text{ mol}} = 77.64 \text{ g/mol}\) Thus, the molar mass of the unknown gas is approximately 77.64 g/mol.

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Most popular questions from this chapter

Perform the following conversions: \((\mathbf{a})\) 0.912 atm to torr, \((\mathbf{b})\) 0.685 bar to kilopascals, \((\mathbf{c})\) 655 \(\mathrm{mm}\) Hg to atmospheres, \((\mathbf{d})\) \(1.323 \times 10^{5}\) Pa to atmospheres, \((\mathbf{e})\) 2.50 atm to psi.

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You have a sample of gas at \(-33^{\circ} \mathrm{C}\) . You wish to increase the rms speed by a factor of \(2 .\) To what temperature should the gas be heated?

An herbicide is found to contain only \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and Cl. The complete combustion of a 100.0 -mg sample of the herbicide in excess oxygen produces 83.16 \(\mathrm{mL}\) of \(\mathrm{CO}_{2}\) and 73.30 \(\mathrm{mL}\) of \(\mathrm{H}_{2} \mathrm{O}\) vapor expressed at STP. A separate analysis shows that the sample also contains 16.44 \(\mathrm{mg}\) of Cl. (a) Determine the percentage of the composition of the substance. ( b) Calculate its empirical formula. (c) What other information would you need to know about this compound to calculate its true molecular formula?

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