Henry-s law constant for molality of methane in benzene at 298K is 4-27-105mm Hg- The mole fraction of methane in benzene at 298K under 760mm Hg is-

The correct option is A 1.78×10^ 3 As per Henry's law #160; p = K_Hx therefore #160; x=pK_H as per the question x= 7604.27× 10^5= 1.78× 10 ...

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Standard XII

Chemistry

Henry's Law

Henry's law c...

Question

Henry's law constant for molality of methane in benzene at $298 K$ is $4.27 \times 10^{5} m m$ $H g$ . The mole fraction of methane in benzene at $298 K$ under $760 m m$ $H g$ is:

1.78 \times 10^{- 3}

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Solution

The correct option is A $1.78 \times 10^{- 3}$
As per Henry's law
$p = K_{H} x$
therefore
$x = \frac{p}{K_{H}}$
as per the question
$x = \frac{760}{4.27 \times 10^{5}} = 1.78 \times 10^{- 3}$

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Similar questions

Q. Henry law constant for the solubility of methane in benzene at 298K is

4.27 \times 10^{5}

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Henry's law constant for molality of methane in benzene at 298K is 4.27×105mm Hg. The mole fraction of methane in benzene at 298K under 760mm Hg is:

The correct option is A 1.78×10−3As per Henry's law p=KHxtherefore x=pKHas per the questionx=7604.27×105=1.78×10−3

Henry's law constant for molality of methane in benzene at $298 K$ is $4.27 \times 10^{5} m m$ $H g$ . The mole fraction of methane in benzene at $298 K$ under $760 m m$ $H g$ is:

The correct option is A $1.78 \times 10^{- 3}$
As per Henry's law
$p = K_{H} x$
therefore
$x = \frac{p}{K_{H}}$
as per the question
$x = \frac{760}{4.27 \times 10^{5}} = 1.78 \times 10^{- 3}$