1.

Calculate the percentage of the naturally occurring isotopes .^(35)Cl and .^(37)Cl that accounts for the atomic mass of chlorine taken as 35.45.

Answer»

Solution :Suppose `.^(35)Cl` present `=x%`. Then `.^(37)Cl` present `=(100-x)%`
`therefore"AVERAGE atomic MASS"=(x xx35+(100-x)xx(37)/(100))=35.45"(Given)"`
`"or"35x+3700-37x=3545 or 2x=155 or x=77.5%`
Thus, `.^(35)Cl=77.5% and .^(37)Cl=100-77.5=22.5%.`


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