Partial pressure of H2 in H2O at 25oC is 1 bar kx for H2 in H2O at 25oC is 71 -18 bar solubility of H2 in H2O will be-

P _ H _ 2 = K _ H X _ H _ 2 X _ H _ 2 = P _ H _ 2 k _ H = 1 bar 7.18× 1 0 ^ 3 bar = 1.404× 1 0 ^ 5 Now, ^ n H _ 2 O in 1L = ...

You visited us 7 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Heat of Reaction

Partial press...

Question

Partial pressure of $H_{2}$ in $H_{2} O$ at $25^{o} C$ is $1$ bar $k^{x}$ for $H_{2}$ in $H_{2} O$ at $25^{o} C$ is $71 \cdot 18$ bar solubility of $H_{2}$ in $H_{2} O$ will be:

Open in App

Solution

$P_{H_{2}} = K_{H} X_{H_{2}}$

$X_{H_{2}} = \frac{P_{H_{2}}}{k_{H}}$ $= \frac{1 b a r}{7.18 \times 1 0^{3} b a r} = 1.404 \times 1 0^{- 5}$

Now, $^{n} H_{2} O i n 1 L = \frac{1000 g}{18} = 55.5$
$\frac{^{n} H_{2}}{55.5} = 1.404 \times 1 0^{- 5}$
$^{n} H_{2} = 1.79 \times 1 0^{- 4} m o l e s$

Conclusion: Hence the solubility of $H_{2}$ in water is $1.79 \times 1 0^{- 4} m o l e s$

Suggest Corrections

Similar questions

Q. Partial

P

H_{2}

H_{2} O

25^{o} C

is 1 bar, constant for

H_{2}

H_{2} O

25^{o} C

71.18

bar. solublity of

H_{2}

H_{2} O

will be ?

Q. Assertion : At

25^{o} C

and 1 atmospheric pressure,

H_{2} O

is a liquid but

H_{2} S

is a gas
Reason : The molecular weight of

H_{2} O

is less than that of

H_{2} S

The standard reduction potential at $25^{o} C$ of the reaction $H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$ is $- 0 .8277 V$ . Calculate the equilibrium constant for the reaction $2 H_{2} O \to H_{3} O^{+} + O H^{-}$ at $25^{o} C$ .

Q. Arrange in decreasing order of boiling point

H_{2} S, H_{2} O, H_{2} S e, H_{2} T e

Q. What will be the oxidation potential for the following hydrogen half cell at

1

atm pressure and

25^{o} C

temperature?

P t ∣ ∣ H_{2 (g)} ∣ ∣ H C l (a q)

when

p H = 3

and pressure of

H_{2}

gas is 1 bar.

Join BYJU'S Learning Program

Partial pressure of H2 in H2O at 25oC is 1 bar kx for H2 in H2O at 25oC is 71⋅18 bar solubility of H2 in H2O will be:

PH2=KHXH2XH2=PH2kH=1bar7.18×103bar=1.404×10−5Now, nH2Oin1L=1000g18=55.5 nH255.5=1.404×10−5 nH2=1.79×10−4molesConclusion: Hence the solubility of H2 in water is 1.79×10−4moles

Partial pressure of $H_{2}$ in $H_{2} O$ at $25^{o} C$ is $1$ bar $k^{x}$ for $H_{2}$ in $H_{2} O$ at $25^{o} C$ is $71 \cdot 18$ bar solubility of $H_{2}$ in $H_{2} O$ will be: