The average atomic mass of a sample of an element X is 16.2 u. What is the % of isotopes ¹⁶ ₈X and ¹⁸ ₈X in the sample?

The sample X is oxygen.

% of O¹⁶₈ - 99.76 %

% of O¹⁸₈ - 20 %

16x+18(100-x)=1620

16x+1800-18x=1620

-2x=-180

x=90

%of 1st isotope of oxygen = 90%

let% of first isotope(mass no.=16) be x and second(mass no.=18) be100-x

Avg. atomic mass=(16× x/100+ 18×100-x/100)

                     16.2    =(16x/100+ 1800-18x/100)

                      16.2    =(16x+1800-18x/100)

                       16.2   =-2x+1800/100

                        16.2×100= -2x+1800, 1620-1800= -2x

                       -2x= -180, x=90, so ist iso. %= x=90%, 2nd iso %= 100-x=100-90= 10%

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It is given that the average atomic mass of the sample of element X is 16.2 u.

Let the percentage of isotope be y%. Thus, the percentage of isotope will be (100 − y) %.

Therefore,

Therefore, the percentage of isotopeis 10%.

And, the percentage of isotopeis (100 − 10) % = 90%.