R.A.M = (Mass.no ×Abundance) (Total abundance)Let the abundance of Li-6 be xRelative abundance of Li-7 will be 100−x.∴ 6.94 = (6 × x)+ 7(100−x) 1006x + 700 – 7x = 694X = 6%Li-6 has 6%, Li-7 has 94%
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