Question

If 4 g of oxygen diffuses through a very narrow hole,how much hydrogen would have diffused under identical conditions?

Answer 1

Hey dear,

● Answer -
1 g H2

● Solution -
Let M1 and M2 be molar masses of O2 and H2 and rate(O2) and rate(H2) be their respective rate of diffusion.

According to Graham's law of diffusion-
rate(O2) / rate(H2) = √(M1 / M2)
4 / rate(H2) = √(32 /2) = 4
rate(H2) = 4/4
rate(H2) = 1 g

Therefore, hydrogen gas diffused through narrow hole under identical conditions is 1 g.

Hope this helps you...

Answer 2

"$Time\quad taken\quad by\quad the\quad 4g\quad of\quad { O }_{ 2 }\quad to\quad diffused\quad =\quad 1$

$And\quad in\quad smae\quad time\quad mass\quad of\quad { H }_{ 2 }\quad diffused\quad /time\quad taken\\ Rate\quad of\quad diffusion\quad of\quad { H }_{ 2 },\quad r{ H }_{ 2 }\quad =\quad \frac { { V }_{ { H }_{ 2 } } }{ t } \quad ...........(1)$

$Rate\quad of\quad diffusion\quad of\quad { O }_{ 2 },\quad r{ O }_{ 2 }\quad =\quad \frac { { V }_{ { H }_{ 2 } } }{ t } ...............(2)$

$On\quad dividing\quad equation\quad (1)\quad and\quad (2)\quad ,\quad$

$We\quad get$

$\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{ 2 } } } =\quad \frac { { V }_{ { H }_{ 2 } } }{ { V }_{ { O }_{ 2 } } }$

$\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{2}}} \quad =\quad \frac { \frac { { m }_{ { H }_{ 2}}}{ { d }_{ { H }_{ 2 }}}}{ \frac { { m }_{ { O }_{2}}}{ { d }_{{O}_{2}}}}$

$\quad \quad \quad \quad \quad =\quad \frac { \frac { { m }_{ { H }_{ 2 } } }{ 0.08986}}{ \frac { 4 }{ 1.429}}$

$\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{ 2 } } } \quad =\quad 3.975\quad \times \quad { m }_{ { H }_{ 2 } }..................(3)\\ According\quad to\quad Graham's\quad law\quad of\quad diffusion$

${ m }_{ { H }_{ 2 } }\quad =\quad 1.006\quad g$

$\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{ 2 } } } \quad =\quad \sqrt { \frac { { M }_{ { O }_{ 2 } } }{ { M }_{ { H }_{ 2 } } }}$

$\frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{ 2 } } } \quad =\quad \sqrt { \frac { 32 }{ 2 } =\quad 4\quad ............(4)$

$Combining\quad equation\quad (3)\quad and\quad (4)$

$\quad \frac { { r }_{ { H }_{ 2 } } }{ { r }_{ { O }_{ 2 } } } \quad =\quad 3.975\quad \times \quad { m }_{ { H }_{ 2}}\quad =\quad 4$

${ m }_{ { H }_{ 2 } }\quad =\quad 1.006\quad g$

Therefore,$\quad the\quad amount\quad of\quad { H }_{ 2 }\quad diffused\quad is\quad 1.006\quad g$"