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Atomic Number and Mass Number

The percentag...

Question

The percentage of oxygen in pure $N a O H$ is (atomic weight of $N a = 23$ ):

40

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16

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8

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23

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Solution

The correct option is A $40$
% of $O = \frac{W e i g h t o f o x y g e n}{M o l e c u l a r w e i g h t o f c o m p o u n d} \times 100$

$= \frac{16}{40} \times 100 = 40$ %

Suggest Corrections

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80 %

N a O H

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100 m l, 1 M H_{2} S O_{4}

N a = 23, H = 1, O = 16, S = 32

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O = 45.3 %

N a = 23, C = 12, O = 16

20

N a = 23 u, O = 16 u

[N a O H]

160

N a = 23, O = 16, H = 1

C

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