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Byju's Answer

Standard XII

Chemistry

Faraday's First Law

Which of the ...

Question

Which of the following correctly represent solubility of $N_{2}$ in water (in terms of $m m o l e / l i t r e$ ) if $N_{2}$ is bubbled through water at $293 K$ at a pressure of $0.768 b a r$ and Henry's constant is $76.8 K$ at $293 K$ ?

10^{- 5}

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5.56 \times 10^{- 4}

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10^{- 2}

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4.3 \times 10^{- 4}

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Solution

The correct option is A $10^{- 5}$
According to Henry's Law
$P_{N_{2}} = K_{M} \times H_{N_{2}}$
$H_{N_{2}} = 0.768 \times 10$
$76.8 \times 10^{2} \times 10^{- 3}$
$= 10^{- 5}$
H_{N_2}=HN_2$
$n$ values
$10^{- 5} = \frac{N O_{2}}{55.55}$
$N O_{2} = 55.55 \times 10^{- 5} m o l e$
Concentration of $N_{2}$
$= 5.56 \times 10^{- 1}$
$= 1555.4 \times 1$
$= 1.0015544$

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Which of the following correctly represent solubility of N2 in water (in terms of mmole/litre) if N2 is bubbled through water at 293 K at a pressure of 0.768 bar and Henry's constant is 76.8 K at 293 K?

The correct option is A 10−5According to Henry's LawPN2=KM×HN2HN2=0.768×1076.8×102×10−3=10−5H_{N_2}=HN_2$n values10−5=NO255.55NO2=55.55×10−5moleConcentration of N2=5.56×10−1=1555.4×1=1.0015544

Which of the following correctly represent solubility of $N_{2}$ in water (in terms of $m m o l e / l i t r e$ ) if $N_{2}$ is bubbled through water at $293 K$ at a pressure of $0.768 b a r$ and Henry's constant is $76.8 K$ at $293 K$ ?