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The ratio of masses of oxyg​en and nitrogen in a particular gaseous mixture 1:4 the ratio of number of their molecule is.

The ratio of masses of oxyg​en and nitrogen in a particular gaseous mixture 1:4 the ratio of number of their molecule is.

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Grade:10

4 Answers

noogler
489 Points
8 years ago
7:32
noogler
489 Points
8 years ago
cuz 1 mole of an atom contains 6.023x1023 molecules(N)
let x be mass of oxygen ….........4x be mass of nitrogen
no. of moles of O2=x/32...............no. of moles of N2=4x/28
no.of molecules of O2=(x/32)N...............no. of molecules of N2=(4x/28)N
then find the ratio of no. of molecules of O2 , N2
Supriya vispute
11 Points
6 years ago
mass of oxygen/mass of nitrogen=1/4. Mole=weight / atomic weightMole of oxygen be 1x/32 and mole of nitrogen be 4x/32=x/7Mole=no.of molecules /NATherefore,mole of oxygen(x/32)=No.of molecules /NA Therefore no.of molecules =NA/32.....1Mole of nitrogen (x/7)=No.of molecules /NATherefore NA/7=no.of molecules .......2From1 and 2NA/32=NA/7Therefore 7NA=32NA.....(cross multiplying) i.e.7:32
pradnya
12 Points
5 years ago
answer: 7:32
iet x be the mass of oxygen
therefore , oxygen=x/32
similarily,
mass of nitrogn is 4x
therefore, nitrogen =4x/28
now the ratio =x/32:4x/28=7:32

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