An element X with relative atomic mass 16.2 contains two isotopes \(^{16}_{8}\)X with relative abundance of 90% and \(^{m}_{8}\)X with relative abundance of 10%. The value of m is
16
18
12
14
Explanation
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solution
16×90/100+m×10/100=16.2
14.4+0.1m=16.2
0.1m=16.2-14.4
0.1m=1.8
0.1m/0.1 = 1.8/0.1
m= 18//
correct answer is option B 
(using their percentage ratio) 90:10= 9:1=(10)
(9/10*16 )= 144/10 =14.4
(1/10*X) =X/10
14.4 + X/10 = 16.2
144+X=162
X = 162-144=18
looking for the relative atomic mass we are going to be adding the two mass the first mass is giving to us as 16 and the second is m we need to look for the value of m
so we simply right it out as an equation
for the first element 16/8X:90/100×16/1...eqn(1)
second element m/8X:10/100×m/1
looking for the relative atomic mass we are going to be adding the two mass the first mass is giving to us as 16 and the second is m we need to look for the value of m
so we simply right it out as an equation
for the first element 16/8X:90/100×16/1...eqn(1)
second element m/8X:10/100×m/1...eqn(2)
normally adding the both gives us our relative atomic mass of 16.2 sow e write it as (90/100×16/1)+(10/100
looking for the relative atomic mass we are going to be adding the two mass the first mass is giving to us as 16 and the second is m we need to look for the value of m
so we simply right it out as an equation
for the first element 16/8X:90/100×16/1...eqn(1)
second element m/8X:10/100×m/1...eqn(2)
normally adding the both gives us our relative atomic mass of 16.2 sow e write it as (90/100×16/1)+(10/100×m/1)=16.2
(144/10)+(m/10)=16.2
the L.C.M is 10
144+m/10=16.2
cross multiply
144+m =16.2×10
144+m=162
you factorizd now by collecting like terms
m=162-144
we all know when a positive is moving to the next it will turn negative
m=18
