4 g CH_4 and 2 g He are mixed under NTP. | vedclass

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(C) Step $1$: Calculate the number of moles of each gas.$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$$n_{He} = \frac{2 \ g}{4 \ g/mol} = 0.5 \ mo

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$\frac{2}{3} \ atm$

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(C) Step $1$: Calculate the number of moles of each gas.$n_{CH_4} = \frac{4 \ g}{16 \ g/mol} = 0.25 \ mol$$n_{He} = \frac{2 \ g}{4 \ g/mol} = 0.5 \ mol$Step $2$: Calculate the total number of moles.$n_{total} = n_{CH_4} + n_{He} = 0.25 + 0.5 = 0.75 \ mol = \frac{3}{4} \ mol$Step $3$: Calculate the mole fraction of $He$.$X_{He} = \frac{n_{He}}{n_{total}} = \frac{0.5}{0.75} = \frac{2}{3}$Step $4$: Calculate the partial pressure of $He$ at $NTP$.At $NTP$,the total pressure $P_{total} = 1 \ atm$.$P_{He} = X_{He} 	imes P_{total} = \frac{2}{3} 	imes 1 \ atm = \frac{2}{3} \ atm$.

$4 \ g \ CH_4$ and $2 \ g \ He$ are mixed under $NTP$. What is the partial pressure of $He$?

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