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Grade: 12th Pass

                        

The density of the mixture of O2 and N2 at STP is 1.3 g/l. Calculate Partial pressure of O2.

7 years ago

Answers : (2)

ankitesh gupta
63 Points
							

CONSIDER A MIXTURE OF N2 AND O2  AT STP DENSITY IS GIVEN TO BE 1.3 g/L CONSIDER THE TOTAL VOLUME TO BE 1L THEN THE TOTAL WEIGHT OF THE MIXTURE WILL BE 1.3 gm SINCE DENSITY=MASS/VOLUME  . NOW CONSIDER THE VOLUME OF N2 BE ''a'' L THEREFORE VOLUME OF O2 WILL BE ''1-a'' L  AND ALSO MASS OF N2 BE ''b''gm THEREFORE MASS OF O2 WILL BE ''1.3-b''gm    NOW AT STP AND NTP MOLES = VOLUME(LITRES)/22.4  THEREFORE

 FOR N2 MOLECULAR MASS IS 28 THEREFORE AND SINCE MOLES = WT/MOLECULAR MASS

 b/28=a/22.4........................(1)

 FOR O2 MOLECULAR MASS IS 32 gm THEREFORE

 (1.3-b)/32=(1-a)/22.4......................(2)

 YOU HAVE TWO EQUATION TWO VARIABLES SOLVE FOR ''a'' AND ''b'' AND NOW THEY ARE ASKIN FOR PARTIAL PRESSURE OF O2

 NOW YOU KNOW THE VOLUME OF O2 AND MASS OF O2

 (PRESSURE OF O2) =(MOLEFRACTION OF O2] * TOTAL PRESSURE.........{TOTAL PRESSURE IS 1 ATM AND MOLEFRACTION = MOLES OF O2/TOTAL MOLES}

TOTAL MOLES CAN BE CALCULATED OCE YOU GET THE VALUE F ''a'' and ''b''...............

HOPE YOU HAVE UNDERSTOOD....................................Laughing 

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7 years ago
naresh narra
18 Points
							

given

N2 and O2 mixture    i.e N2+O2      density = 1.3g/l at stp

 

i.e     1 lit = 1.3g

therefore   22.4 lit = 22.4 * 1.3

     = 29.12 g

we know that at 22.4 lit the weight is known as molecular weight

therfore mw(M)= 29.12 g

wkt.,n=w/M

      w= n*M

for given eq., N2 +O2

   total weight at 22.4 lit (stp) is = 28x + 32 (1-x)

=>   29.12  =   28x+32-32x

therefore   x = 0.72

therefore  no.of moles of N2= 0.72 and O2 = 1- 0.72 = 0.28

therefore partial pressure of O2 at stp PO2 = n O2/(n O2 + n N2 )  *  p total

                                                         p O2    =   0.28/(0.28+0.72)   *    1atm

                                                                     p  O2 =    0.28 atm.

7 years ago
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