Find the percentage of oxygen in magnesium nitrate crystals, $M g (N O_{3})_{2} .6 H_{2} O$ [Mg = 24, N = 14, O = 16, H = 1]

70%

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75%

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60%

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80%

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Solution

The correct option is
D

75%

×
Molecular Mass of $M g (N O_{3})_{2} .6 H_{2} O$ = 24 + (2 $\times$ 62) + 6 $\times$ 18 = 24 + 108 = 256 g

Mass of Oxygen in $M g (N O_{3})_{2} .6 H_{2} O$ = 12 ×16 = 192 g

Percentage of Oxygen = $\frac{192}{256} \times$ 100 = 75%

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Similar questions

Q. What is the total percentage of oxygen in magnesium nitrate crystals

(M g (N O_{3})_{2} .6 H_{2} O)

[

O = 16, N = 14, H = 1, M g = 24

]

Q. The total percentage of magnesium present in magnesium nitrate crystals

M g (N O_{3})_{2} .6 H_{2} O

is:
[

M g = 24; N = 14; O = 16; H = 1

Find the total percentage of oxygen in magnesium nitrate crystal Mg(NO₃)₂·6H₂O.
[H = 1, N = 14, O = 16, Mg = 24]

Q. Find the total percentage of Magnesium in magnesium nitrate crystals:

[M g (N O_{3})_{2} 6 H_{2} O .

(M g = 24, O = 16

and

H = 1)

Find the percentage of
(a) oxygen in magnesium nitrate crystals $[M g (N O_{3})_{2} 6 H_{2} O]$ .
(b) boron in $N a_{2} B_{4} O_{7} . 10 H_{2} O . [H = 1, B = 11, O = 16, N a = 23] .$
(c) phosphorus in the fertilizer superphosphate $C a (H_{2} P O_{4})_{2}$

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Find the percentage of oxygen in magnesium nitrate crystals, Mg(NO3)2.6H2O [Mg = 24, N = 14, O = 16, H = 1]

The correct option is D 75%×Molecular Mass of Mg(NO3)2.6H2O = 24 + (2 × 62) + 6 × 18 = 24 + 108 = 256 g Mass of Oxygen in Mg(NO3)2.6H2O = 12 ×16 = 192 g Percentage of Oxygen = 192256× 100 = 75%

Find the percentage of oxygen in magnesium nitrate crystals, $M g (N O_{3})_{2} .6 H_{2} O$ [Mg = 24, N = 14, O = 16, H = 1]

The correct option is
D

75%

×
Molecular Mass of $M g (N O_{3})_{2} .6 H_{2} O$ = 24 + (2 $\times$ 62) + 6 $\times$ 18 = 24 + 108 = 256 g

Mass of Oxygen in $M g (N O_{3})_{2} .6 H_{2} O$ = 12 ×16 = 192 g

Percentage of Oxygen = $\frac{192}{256} \times$ 100 = 75%