QuestionMay 13, 2025

There are 103.0 g of chlorine gas (Cl_(2)) and 23.0 g of helium (He) mixed in a container with a total pressure of 1,741 mmHg. Determine the partial pressure of helium gas.Round your answer to the nearest hundredths place. Answer: underline ( )mmHg square

There are 103.0 g of chlorine gas (Cl_(2)) and 23.0 g of helium (He) mixed in a container with a total pressure of 1,741 mmHg. Determine the partial pressure of helium gas.Round your answer to the nearest hundredths place. Answer: underline ( )mmHg square
There are 103.0 g of chlorine gas (Cl_(2)) and 23.0 g of helium (He) mixed in a container with a total pressure of 1,741 mmHg. Determine the partial pressure of helium gas.Round your answer to the nearest hundredths place. Answer: underline ( )mmHg square

Solution
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Answer

1389.32 mmHg Explanation 1. Calculate moles of each gas Moles of Cl_2: \frac{103.0 \, \text{g}}{70.9 \, \text{g/mol}} = 1.452 \, \text{mol}; Moles of He: \frac{23.0 \, \text{g}}{4.0 \, \text{g/mol}} = 5.75 \, \text{mol}. 2. Calculate total moles Total moles = 1.452 + 5.75 = 7.202 \, \text{mol}. 3. Calculate mole fraction of helium Mole fraction of He = \frac{5.75}{7.202} = 0.798. 4. Calculate partial pressure of helium **Partial Pressure Formula**: P_{\text{He}} = X_{\text{He}} \times P_{\text{total}}; P_{\text{He}} = 0.798 \times 1741 \, \text{mmHg} = 1389.318 \, \text{mmHg}.

Explanation

1. Calculate moles of each gas<br /> Moles of $Cl_2$: $\frac{103.0 \, \text{g}}{70.9 \, \text{g/mol}} = 1.452 \, \text{mol}$; Moles of He: $\frac{23.0 \, \text{g}}{4.0 \, \text{g/mol}} = 5.75 \, \text{mol}$.<br /><br />2. Calculate total moles<br /> Total moles = $1.452 + 5.75 = 7.202 \, \text{mol}$.<br /><br />3. Calculate mole fraction of helium<br /> Mole fraction of He = $\frac{5.75}{7.202} = 0.798$.<br /><br />4. Calculate partial pressure of helium<br /> **Partial Pressure Formula**: $P_{\text{He}} = X_{\text{He}} \times P_{\text{total}}$; $P_{\text{He}} = 0.798 \times 1741 \, \text{mmHg} = 1389.318 \, \text{mmHg}$.
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