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I m not able to solve this question. Can u please answer with detailed solution: The density of a mixture of O 2 and N 2 at STP is 1.3g/litre. Calculate the partial pressure of O 2.
hi, let the total no. of moles be n in which n1 are of O2 & n2 are of N2 so, n= n1 + n2 let the volume be V , so density = total mass/ V 1.3 = (n1 *32 + n2 *28 )/V ...................1 at stp , Volume of one mole gas is 22.4 lt. so the volume of n= n1 +n2 moles = 22.4 n V = 22.4 (n1 + n2 ) putting in eq 1 1.3 = (n1 *32 + n2 *28 )/ 22.4 (n1 + n2 ) 29.12 n1 + 29.12 n2 = 32 n1 + 28 n2 1.12 n2 = 2.88 n1 n1/n2 = .389 so, n1/n = .2801 ( n2/n1 + 1 = (n1 +n2) /n1) at stp total P = 101.325 k Pa partial pressure of O2 = n1/n* P = .2801 *101.325 = 28.37 k Pa
hi,
let the total no. of moles be n in which n1 are of O2 & n2 are of N2
so, n= n1 + n2
let the volume be V ,
so density = total mass/ V
1.3 = (n1 *32 + n2 *28 )/V ...................1
at stp , Volume of one mole gas is 22.4 lt.
so the volume of n= n1 +n2 moles = 22.4 n
V = 22.4 (n1 + n2 ) putting in eq 1
1.3 = (n1 *32 + n2 *28 )/ 22.4 (n1 + n2 )
29.12 n1 + 29.12 n2 = 32 n1 + 28 n2
1.12 n2 = 2.88 n1
n1/n2 = .389
so, n1/n = .2801 ( n2/n1 + 1 = (n1 +n2) /n1)
at stp total P = 101.325 k Pa
partial pressure of O2 = n1/n* P
= .2801 *101.325
= 28.37 k Pa
Thank You!! so much for the reply..
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