KCET 2025 Chemistry Paper

To determine the number of moles of N2 gas that would dissolve in 1 liter of water at 293 K , when N2 exerts a partial pressure of 0.987 bar, we can use Henry's Law. Given that the Henry's Law constant (KH) for N2 at 293 K is 76.48 K bar, we apply the following:
Henry's Law
P=KH⋅X
Where:
P is the partial pressure of the gas ( 0.987 bar ),
KH is Henry's Law constant ( 76.48 K bar),
X is the mole fraction of the gas in the solvent.
Calculating Mole Fraction of N2
The mole fraction XN2 is given by:
XN2=‌

PN2

KH

=‌

0.987

76.48×103

=1.29×10−5
Determining Moles of Water
The number of moles of water ( nH2O ) in 1 liter ( 1000 grams) is:
nH2O=‌

1000

18

=55.5
Relating Mole Fraction and Moles of N2
The mole fraction can also be expressed as:
XN2=‌

nN2

nN2+nH2O

≈‌

nN2

55.5

Setting this equal to the previously calculated mole fraction:
1.29×10−5=‌

n

55.5

Solving for n (the moles of N2 ):
nN2=1.29×10−5×55.5=7.16×10−4
Thus, the number of moles of N2 gas dissolved in 1 liter of water is 7.16×10−4.