The value of Henry's law constant for some gases at $293 K$ is given below. Arrange the gases in the increasing order of their solubility. $H e = 144.97 k b a r$ ; $H_{2} = 69.16 k b a r$
$N_{2} = 76.48 k b a r$ ; $O_{2} = 34.86 k b a r$ :

H e < N_{2} < H_{2} < O_{2}

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O_{2} < H_{2} < N_{2} < H e

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H_{2} < N_{2} < O_{2} < H e

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H e < O_{2} < N_{2} < H_{2}

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Solution

The correct option is A $H e < N_{2} < H_{2} < O_{2}$
$K_{H} = \frac{p}{x}$

At constant pressure, $K_{H}$ is inversely proportional to $x$ . Therefore, the higher the $K_{H}$ lower the solubility.
The increasing solubility order is:
$H e < N_{2} < H_{2} < O_{2}$

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Similar questions

Henrys law constant $[K_{H}]$ for four gases at 293 K are,

$H e [144.97 K b a r], H_{2} [69.16 K b a r], N_{2} [76.48 K b a r], O_{2} [34.86 K b a r]$ .

Which gas is more soluble in water at a given pressure.

Gases posses charcteristic critical temperature which depends upo the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

$\begin{matrix} G a s e s & H_{2} & H e & O_{2} & N_{2} C r i t i c a l t e m p e r a t u r e i n K e l v i n & 33.2 & 5.3 & 154.3 & 126 \end{matrix}$

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first

(a) $H_{2}, H e, O_{2}, N_{2}$ (b) $H e, O_{2}, H_{2}, N_{2}$

Q. Gases

H_{2} H e O_{2} N_{2}

Critical temperature(K) 33.2 5.3 154.3 126

From the above data what whould be the order of liquefaction of these gases?
Start writing the order from the gas liquefying first.

Q. At

S T P

the order of mean square velocity of molecules of

H_{2}, N_{2}, O_{2}

and

H B r

is-

Q. State whether true or false:

Mean free path order for some gases at 273 K and 1 atm P is

H e > H_{2} > O_{2} > N_{2} > C O_{2}

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The value of Henry's law constant for some gases at 293K is given below. Arrange the gases in the increasing order of their solubility. He=144.97kbar; H2=69.16kbarN2=76.48kbar; O2=34.86kbar:

The correct option is A He<N2<H2<O2KH=pxAt constant pressure, KH is inversely proportional to x. Therefore, the higher the KH lower the solubility.The increasing solubility order is:He<N2<H2<O2

The value of Henry's law constant for some gases at $293 K$ is given below. Arrange the gases in the increasing order of their solubility. $H e = 144.97 k b a r$ ; $H_{2} = 69.16 k b a r$
$N_{2} = 76.48 k b a r$ ; $O_{2} = 34.86 k b a r$ :

The correct option is A $H e < N_{2} < H_{2} < O_{2}$
$K_{H} = \frac{p}{x}$

At constant pressure, $K_{H}$ is inversely proportional to $x$ . Therefore, the higher the $K_{H}$ lower the solubility.
The increasing solubility order is:
$H e < N_{2} < H_{2} < O_{2}$