Problem 12 - Chemistry in 30 days

Question

The density in grams per litre of a mixture containing an equal number of moles of methane and ethane at STP is (a) \(1.03\) (b) \(1.10\) (c) \(0.94\) (d) \(1.20\)

Step-by-Step Solution

Answer

The density is approximately 1.03 g/L, matching option (a).

Step 1: Determine Molar Masses

Calculate the molar mass of methane \(CH_4\) which is \(12 + 4 = 16\) grams/mole. Similarly, calculate the molar mass of ethane \(C_2H_6\) which is \(24 + 6 = 30\) grams/mole.

Step 2: Understand STP Conditions

At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 liters.

Step 3: Calculate Total Molar Mass

Because the mixture contains an equal number of moles of methane and ethane, average the molar masses: \((16 + 30)/2 = 23\) grams/mole.

Step 4: Calculate Density

Using the volume of 22.4 L/mol at STP, calculate the density: \(23 \, \text{grams/mole} / 22.4 \, \text{L/mole} = 1.0268 \, \text{grams/L}\).

Step 5: Round Off to Closest Option

Round 1.0268 to two decimal places, which becomes 1.03. Hence, the correct answer from the options given is (a) 1.03.

Key Concepts

Understanding Molar Mass

Molar mass is a way to express how much one mole of a substance weighs. It's like the weight of a dozen apples, just tailored for molecules. Each element has its own molar mass, and you can find this information on the periodic table as it represents the mass of one mole of that element in grams. For compounds, you calculate molar mass by adding the molar masses of each atom in the compound.

• For methane (\(CH_4\)), it's 12 grams for carbon and 1 gram for each of the 4 hydrogens: \(12 + 4 = 16\) grams/mole.
• For ethane (\(C_2H_6\)), it's 24 for two carbons and 6 for the hydrogens: \(24 + 6 = 30\) grams/mole.

Knowing molar mass is crucial for various calculations in chemistry, such as converting moles to grams or finding the density of gases.

STP Conditions Explained

STP stands for Standard Temperature and Pressure. It's a set of conditions used for scientific calculations when dealing with gases, to ensure consistency. At STP, the temperature is standardized to 0°C (273.15 K) and the pressure is atmospheric pressure, or 1 atm.
One of the nifty things about STP is that one mole of any ideal gas occupies 22.4 liters. This simplifies calculations because:

• It gives you a common baseline for comparing how different gases behave under the same conditions.
• Makes it easier to calculate properties like density.

Thus, knowing that gases occupy this volume at STP helps make density calculations straightforward.

Ideal Gas Law Insights

The Ideal Gas Law is a handy equation that relates the pressure, volume, temperature, and number of moles of a gas. It's written as \(PV = nRT\), where:

• \(P\) is pressure (in atm)
• \(V\) is volume (in liters)
• \(n\) is the number of moles
• \(R\) is the ideal gas constant (0.0821 \(L \, atm \, K^{-1} \, mol^{-1}\))
• \(T\) is temperature (in Kelvin)

The Ideal Gas Law makes it easy to determine one property if the others are known. At STP, you can use these conditions and the molar volume (22.4 liters/mole) to solve directly for properties like density.

Density Calculation in Gases

Density is a measure of how much mass is contained in a given volume. For gases, calculating density can seem tricky, but it's quite simple when you break it down. The formula is: \[ \text{Density} \; (\rho) = \frac{M \; (\text{grams/mole})}{V \; (\text{liters/mole at STP})} \] where \(M\) is the molar mass and \(V\) is 22.4 L (at STP).

• In our exercise, for the mixture of methane and ethane:
  Combine their molar masses because they are mixed equally: \( (16 + 30)/2 = 23 \; \text{grams/mole} \)
• Then, divide by the molar volume: \(23/22.4 = 1.0268 \; \text{grams/L} \)

This calculation shows how you can arrive at the density of a gas mixture and why the result is closer to option (a)