<h1>Problem</h1>
A gas mixture at 25°C and 1 atm contains 100 mg/L of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>H</mi><mn>2</mn></msub><mi>S</mi></mrow><annotation encoding="application/x-tex">H_2S</annotation></semantics></math>H2​S. What is the partial pressure exerted by <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>H</mi><mn>2</mn></msub><mi>S</mi></mrow><annotation encoding="application/x-tex">H_2S</annotation></semantics></math>H2​S. Round answer to two significant digits.
<h1>Solution</h1>
The molecular weight of <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>H</mi><mn>2</mn></msub><mi>S</mi></mrow><annotation encoding="application/x-tex">H_2S</annotation></semantics></math>H2​S is approximately 34.07 g/mol.
We can use the molecular weight to find out how many moles of hydrogen sulfide are in our mixture:
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mrow><mn>100</mn><mtext> mg</mtext></mrow><mtext>L</mtext></mfrac><mo>∗</mo><mfrac><mrow><mn>1</mn><mtext> g</mtext></mrow><mrow><mn>1000</mn><mtext> mg</mtext></mrow></mfrac><mo>∗</mo><mfrac><mtext> mol</mtext><mrow><mn>34.07</mn><mtext> g</mtext></mrow></mfrac><mo>=</mo><mn>0.002935</mn><mtext> mol </mtext><msub><mi>H</mi><mn>2</mn></msub><mi>S</mi></mrow><annotation encoding="application/x-tex">\frac{100 \text{ mg}}{\text{L}} * \frac{1 \text{ g}}{1000 \text{ mg}} * \frac{\text{ mol}}{34.07 \text{ g}} = 0.002935 \text{ mol }H_2S</annotation></semantics></math>L100 mg​∗1000 mg1 g​∗34.07 g mol​=0.002935 mol H2​S</div>
We can then use the ideal gas law to find the pressure:
<div class="math math-display"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>P</mi><mi>V</mi><mo>=</mo><mi>n</mi><mi>R</mi><mi>T</mi><mo>⇒</mo><mi>P</mi><mo>=</mo><mfrac><mrow><mi>n</mi><mi>R</mi><mi>T</mi></mrow><mi>V</mi></mfrac><mo>=</mo><mfrac><mrow><mo stretchy="false">(</mo><mn>0.002935</mn><mtext> mol</mtext><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>0.0821</mn><mfrac><mtext>atm*L</mtext><mtext>mol * K</mtext></mfrac><mo stretchy="false">)</mo><mo stretchy="false">(</mo><mn>298</mn><mi>K</mi><mo stretchy="false">)</mo></mrow><mrow><mo stretchy="false">(</mo><mn>1</mn><mtext> L</mtext><mo stretchy="false">)</mo></mrow></mfrac><mo>≃</mo><mn>0.072</mn><mtext> atm</mtext></mrow><annotation encoding="application/x-tex">PV = nRT \Rightarrow P = \frac{nRT}{V} = \frac{(0.002935 \text{ mol})(0.0821 \frac{\text{atm*L}}{\text{mol * K}})(298 K)}{(1 \text{ L})} \simeq 0.072 \text{ atm}</annotation></semantics></math>PV=nRT⇒P=VnRT​=(1 L)(0.002935 mol)(0.0821mol * Katm*L​)(298K)​≃0.072 atm</div>

Partial Pressure in a Gas Mixture