According to Graham-s Law- the rate of diffusion of CO - O 2- N 2 and CO 2 follows the order-A- CO - N 2- O 2- CO 2B- CO - N 2- CO 2- O 2C- O 2- CO - N 2- CO 2D- CO 2- O 2- CO - N 2

The correct option is A CO = N_2 gt; O_2 gt; CO_2According to Graham's law of effusion, r ∝1√(M) nbsp; Rate of diffusion decrease with increase in mole ...

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Byju's Answer

Standard XII

Chemistry

Dalton's Law of Partial Pressures

According to ...

Question

According to Graham's Law, the rate of diffusion of $C O, O_{2}, N_{2} a n d C O_{2}$ follows the order:

C O = N_{2} > O_{2} > C O_{2}

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C O = N_{2} > C O_{2} > O_{2}

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O_{2} > C O = N_{2} > C O_{2}

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C O_{2} > O_{2} > C O = N_{2}

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Solution

The correct option is A $C O = N_{2} > O_{2} > C O_{2}$
According to Graham's law of effusion,
$r \propto \frac{1}{\sqrt{M}}$
Rate of diffusion decrease with increase in molecular weight.
$∴$ Order of rate of effusion is:
$C O = N_{2} > O_{2} > C O_{2}$
( 28 28 32 44)

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Similar questions

Q. According to Graham's Law, the rate of diffusion of

C O, O_{2}, N_{2}

and

C O_{2}

follows the order.

In which of the following equilibria, the value of $K_{P}$ is less than $K_{C}$

Q. Only

N_{2}

, CO and

C O_{2}

gases remain after 0.72 gm of carbon is treated with one litre of air at

27^{0} C

and 4.92 atm pressure. Assume air composition

O_{2}

= 20%,

N_{2}

= 79% and

C O_{2}

= 1% (by volume). The heat evolved (in Kcal) under constant pressure is:
Given :

C + O_{2} \to C O_{2}, Δ H = - 100 K c a l / m o l

C + \frac{1}{2} O_{2} \to C O, Δ H = - 25 K c a l / m o l

Q. In the reaction;

C O + \frac{1}{2} O_{2} \to C O_{2}; N_{2} + O_{2} \to 2 N O

10 mL of mixture containing carbon monoxide and nitrogen required 7 mL oxygen to form

C O_{2}

and NO, on combustion.
The volume of

N_{2}

in the mixture will be:

C O_{2}, C H_{4}, C O, N H_{3}, water vapour, N_{2} O, O_{2}, S O_{2}, N_{2}, hydrofluorocarbons, O_{3}, C F C

The number of green house gases from the above group is:

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According to Graham's Law, the rate of diffusion of CO,O2,N2 and CO2 follows the order:

The correct option is A CO=N2>O2>CO2According to Graham's law of effusion, r∝1√M Rate of diffusion decrease with increase in molecular weight. ∴ Order of rate of effusion is: CO=N2>O2>CO2 ( 28 28 32 44)

According to Graham's Law, the rate of diffusion of $C O, O_{2}, N_{2} a n d C O_{2}$ follows the order:

The correct option is A $C O = N_{2} > O_{2} > C O_{2}$
According to Graham's law of effusion,
$r \propto \frac{1}{\sqrt{M}}$
Rate of diffusion decrease with increase in molecular weight.
$∴$ Order of rate of effusion is:
$C O = N_{2} > O_{2} > C O_{2}$
( 28 28 32 44)