At 27^{circ} C in a 10 L flask,4.0 g of an | vedclass

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(B) Given:$T = 27 + 273 = 300 \ K$$V = 10 \ L$$Molar mass (He) = 4 \ g \ mol^{-1}$$Molar mass (Ne) = 20 \ g \ mol^{-1}$$P = 1.23 \ atm$$R = 0.082 \ L

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$62.5$

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(B) Given:$T = 27 + 273 = 300 \ K$$V = 10 \ L$$Molar mass (He) = 4 \ g \ mol^{-1}$$Molar mass (Ne) = 20 \ g \ mol^{-1}$$P = 1.23 \ atm$$R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1}$Using the ideal gas equation $PV = nRT$,where $n = n_{He} + n_{Ne} = \frac{W_{He}}{M_{He}} + \frac{W_{Ne}}{M_{Ne}}$:$PV = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20})RT$$1.23 	imes 10 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) 	imes 0.082 	imes 300$$12.3 = (\frac{W_{He}}{4} + \frac{W_{Ne}}{20}) 	imes 24.6$$\frac{W_{He}}{4} + \frac{W_{Ne}}{20} = \frac{12.3}{24.6} = 0.5$Multiplying by $20$:$5W_{He} + W_{Ne} = 10$ (Equation $i$)We are given the total mass:$W_{He} + W_{Ne} = 4$ (Equation $ii$)Subtracting Equation $ii$ from Equation $i$:$(5W_{He} + W_{Ne}) - (W_{He} + W_{Ne}) = 10 - 4$$4W_{He} = 6 \Rightarrow W_{He} = 1.5 \ g$$W_{Ne} = 4 - 1.5 = 2.5 \ g$Mass $\%$ of neon $= \frac{W_{Ne}}{	ext{Total mass}} 	imes 100 = \frac{2.5}{4} 	imes 100 = 62.5 \%$

At $27^{\circ} C$ in a $10 \ L$ flask,$4.0 \ g$ of an ideal gaseous mixture containing $He$ (molar mass $4.0 \ g \ mol^{-1}$) and $Ne$ (molar mass $20 \ g \ mol^{-1}$) has a pressure of $1.23 \ atm$. What is the mass $\%$ of neon? $(R = 0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

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