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Standard XII

Chemistry

Henry's Law

If N2 gas b...

Question

If $N_{2}$ gas bubbled through water at $293 K$ how many millimoles of $N_{2}$ gas would dissolve in 1 litre of water? Assume that exerts a partial pressure of $0.987 b a r$ .
Given that Henry's law constant for $N_{2}$ at $293 K$ is $76.48 K b a r$ :

0.716

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Solution

The correct option is A $0.716$
By Henry's Law, $K_{n} . x = p \Rightarrow K_{n} . \frac{m o l e s}{\frac{1000}{18}} = p$
$\Rightarrow m o l e s = \frac{0.987}{76.48 \times 10} \times \frac{10}{18}$

$\Rightarrow m o l e s = 0.716 \times 10^{- 3} m o l e s$

$= 0.716 m m o l e s$

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Similar questions

Q. If

N_{2}

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293 K

, Calculate solubility of

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^{N₂gas is bubbled through water at 298K. How many moles of N₂ gas would dissolve in 1L water assumed that partial pressure of gas is = 0.987 bar and K_H = 76.48 K .}

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If N2 gas bubbled through water at 293K how many millimoles of N2 gas would dissolve in 1 litre of water? Assume that exerts a partial pressure of 0.987bar. Given that Henry's law constant for N2 at 293K is 76.48Kbar :

The correct option is A 0.716By Henry's Law,Kn.x=p⇒Kn.moles100018=p⇒moles=0.98776.48×10×1018⇒moles=0.716×10−3moles=0.716 mmoles

If $N_{2}$ gas bubbled through water at $293 K$ how many millimoles of $N_{2}$ gas would dissolve in 1 litre of water? Assume that exerts a partial pressure of $0.987 b a r$ .
Given that Henry's law constant for $N_{2}$ at $293 K$ is $76.48 K b a r$ :

The correct option is A $0.716$
By Henry's Law, $K_{n} . x = p \Rightarrow K_{n} . \frac{m o l e s}{\frac{1000}{18}} = p$
$\Rightarrow m o l e s = \frac{0.987}{76.48 \times 10} \times \frac{10}{18}$

$\Rightarrow m o l e s = 0.716 \times 10^{- 3} m o l e s$

$= 0.716 m m o l e s$