Math  /  Algebra

QuestionWhat is the molar mass of a gas with a density of 0.657 g/L0.657 \mathrm{~g} / \mathrm{L} at 1.00 atm and 298 K ? R=0.08206 L atm/mol K\mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} / \mathrm{mol} \mathrm{~K}

Studdy Solution

STEP 1

1. The gas behaves ideally.
2. The density of the gas is 0.657g/L0.657 \, \mathrm{g/L}.
3. The pressure is 1.00atm1.00 \, \mathrm{atm}.
4. The temperature is 298K298 \, \mathrm{K}.
5. The ideal gas constant RR is 0.08206Latm/molK0.08206 \, \mathrm{L} \, \mathrm{atm} / \mathrm{mol} \, \mathrm{K}.
6. We are trying to find the molar mass of the gas.

STEP 2

1. Use the ideal gas law to relate the given quantities.
2. Rearrange the ideal gas law to solve for molar mass.
3. Substitute the known values into the equation.
4. Calculate the molar mass.

STEP 3

Use the ideal gas law, which is given by the equation:
PV=nRT PV = nRT
where P P is the pressure, V V is the volume, n n is the number of moles, R R is the ideal gas constant, and T T is the temperature.

STEP 4

Rearrange the ideal gas law to solve for molar mass (M M ). We know that n=mM n = \frac{m}{M} , where m m is the mass of the gas. Substitute this into the ideal gas law:
PV=mMRT PV = \frac{m}{M}RT
Rearrange to solve for M M :
M=mRTPV M = \frac{mRT}{PV}

STEP 5

Substitute the known values into the equation. We know that density (d d ) is given by d=mV d = \frac{m}{V} , so m=dV m = dV . Substitute m=dV m = dV into the equation for molar mass:
M=dVRTPV M = \frac{dVRT}{PV}
Since V V cancels out, we have:
M=dRTP M = \frac{dRT}{P}
Substitute the known values:
M=0.657g/L×0.08206Latm/molK×298K1.00atm M = \frac{0.657 \, \mathrm{g/L} \times 0.08206 \, \mathrm{L} \, \mathrm{atm/mol} \, \mathrm{K} \times 298 \, \mathrm{K}}{1.00 \, \mathrm{atm}}

STEP 6

Calculate the molar mass:
M=0.657×0.08206×2981.00 M = \frac{0.657 \times 0.08206 \times 298}{1.00}
M=16.00g/mol M = 16.00 \, \mathrm{g/mol}
The molar mass of the gas is:
16.00g/mol \boxed{16.00 \, \mathrm{g/mol}}

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