chlorine occurs in nature in 2 isotropic forms with masses 35u and 37u in the ratio of 3:1. Then what should we take the mass of chlorine atom and why?
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So komli, for any such question where you have to determine mass of final element, you just need to multiple the ratio to corresponding occurrence and divide it by sum of the ratios....it's an averaging mechanism......
Let's consider your question:
Here Cl 35u : 37u is 3 : 1
So, solution becomes:
mass of Chlorine = ( 35*3 + 37*1 ) / ( 3+1 )
= (105 + 37)/4
=35.5
So, final answer becomes 35.5u.
Hope it helps....feel free to ask your doubts.
Let's consider your question:
Here Cl 35u : 37u is 3 : 1
So, solution becomes:
mass of Chlorine = ( 35*3 + 37*1 ) / ( 3+1 )
= (105 + 37)/4
=35.5
So, final answer becomes 35.5u.
Hope it helps....feel free to ask your doubts.
komli:
thank you so much
My pleasure ..komli.
okay
mujhe ye ques likhna h to m isse kha se likhu mtlab kha se start kru likhna
aap start kriye us line se jahan pe likha hai...Cl 35 : 37
ok thnks
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