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Question
Find the percentage of boron in Na2B4O7.10H2O. [H = 1, B = 11, O = 16, Na = 23].
Solution
The molecular mass of Na2B4O7.10H2O = 2(Na) + 4(B) + 7(O) + 10[2 + 16]
= 2(23) + 4(11) + 7(16) + 180
= 46 + 44 + 112 + 180
= 382
Percentage of B in (Na2B4O7.10H2O) = `44/382 xx 100`
= 11.5%
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