+91-87964 74404
FlagBACK
Question icon
Grade 12IIT JEE Entrance Exam

How much oxygen is dissolved in 100mL water at 298K if partial pressure
of oxygen is 0.5 atm and Kh=1.4*10-3mol/L/atm?

Profile image of jakira
9 Years agoGrade 12
Answers icon

1 Answer

Profile image of Nida
7 Years ago
Concentration = partial pressure ×kH
                      n/V= p ×kH
             n= p×kH× V
 Put values, n= 0.5×1.4×10-3 × 0.1
              n=0.00007
32g of  Oxygen = 0.00007×32
                          = 2.24×10-3
                        =2.24mg

Other Related Questions on iit jee entrance exam

question mark

Let S={x:x belongs to R(√3+√2)^(x^2-4) - (√3-√2)^(x^2-4)=10}. Then n(S) is equal to-

iit jee entrance exam
1 Answer Available

Last Activity: 10 Months ago

question mark
My rank is 29k, General, should I opt for any state college cse branch or some Nits in which in which I am getting ECE , chemical, mechanics etc....
 
iit jee entrance exam
1 Answer Available

Last Activity: 3 Years ago

question mark
How is Ranquer Eduzone,comparing, Akash,Fitjee,Resonance,Naryana
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago

question mark
Can class 9th students study syllabus class  of 11th 
iit jee entrance exam
3 Answers Available

Last Activity: 4 Years ago

question mark
Sum of series 2[n/2] ! [n/2]!/ n! [2Co-2C2+3C2....+[-1]^n[n+1]2Cn] where n is an even positive integer. Kindly provide solution of this question and how to generalise this. How do we know that which expansion will give this?
iit jee entrance exam
1 Answer Available

Last Activity: 4 Years ago