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Standard X

Chemistry

Gay Lussac's Law

It takes 26 s...

Question

It takes 26 s for 10 mL of $H_{2}$ to effuse through a porous membrane. It takes 130 s for 10 mL of an unknown gas to effuse. Hence, molecular weight of the unknown gas is (in $g m o l^{- 1}$ )

100

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Solution

The correct option is C
50

$r_{H_{2}} = \frac{10}{26} m L s^{- 1}$

$r_{x} = \frac{10}{130} m L s^{- 1}$

$\frac{r_{H_{2}}}{r_{x}} = \frac{\frac{10}{26}}{\frac{10}{130}} = \sqrt{\frac{m_{x}}{m_{H_{2}}}} = \sqrt{\frac{m_{x}}{2}}$

$5 = \sqrt{\frac{m_{x}}{2}}$

$∴ m_{x} = 50 g m o l^{- 1}$

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It takes 26 s for 10 mL of H2 to effuse through a porous membrane. It takes 130 s for 10 mL of an unknown gas to effuse. Hence, molecular weight of the unknown gas is (in gmol−1 )

The correct option is C 50 rH2=1026 mLs−1 rx=10130 mLs−1 rH2rx=102610130=√mxmH2=√mx2 5=√mx2 ∴mx=50 gmol−1

It takes 26 s for 10 mL of $H_{2}$ to effuse through a porous membrane. It takes 130 s for 10 mL of an unknown gas to effuse. Hence, molecular weight of the unknown gas is (in $g m o l^{- 1}$ )

The correct option is C
50

$r_{H_{2}} = \frac{10}{26} m L s^{- 1}$

$r_{x} = \frac{10}{130} m L s^{- 1}$

$\frac{r_{H_{2}}}{r_{x}} = \frac{\frac{10}{26}}{\frac{10}{130}} = \sqrt{\frac{m_{x}}{m_{H_{2}}}} = \sqrt{\frac{m_{x}}{2}}$

$5 = \sqrt{\frac{m_{x}}{2}}$

$∴ m_{x} = 50 g m o l^{- 1}$