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Question

N​​​​​​2 gas is bubbled through water at 298K. How many moles of N​​​​​​2 gas would dissolve in 1L water assumed that  partial pressure of gas is = 0.987 bar and K​​​​​​H = 76.48 K .


Solution

KH = 76.48kbar = 76480bar
 
P= 0.989bar
 
P = KH × x
x= 0.98976480=1.293 ×10−5. This is the mole fraction of N2 gas dissolved in water

Volume of water =  1L = 1000mL
Taking density as 1g/mL,  mass of 1L =1000g
Moles of water in 1L =  1000/18=55.56
Let us take the moles of nitrogen as x
Total moles = moles of N2 + moles of water = x + 55.56
mole fraction of N2
 1.293 ×10−5= xx+55.561.293 ×10-5x + 1.293 ×10-5 × 55.56 =x x(1-1.293 ×10-5) = 1.293 ×10-5 × 55.56 x=7.17 ×10-4mol = 0.717mmoles

Chemistry

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