The total percentage of magnesium present in magnesium nitrate crystals $M g (N O_{3})_{2} .6 H_{2} O$ is:
[ $M g = 24; N = 14; O = 16; H = 1$ ].

9.38%

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18.76%

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11.93%

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27.43%

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Solution

The correct option is A 9.38%
Molar mass of $M g (N O_{3})_{2} .6 H_{2} O$ , $= 256 g m$

So, % of $M g$ $= 24 \times \frac{100}{256.4} = 9.38$ %

Hence, option $A$ is correct.

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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

]

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]

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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The total percentage of magnesium present in magnesium nitrate crystals Mg(NO3)2.6H2O is:[Mg=24;N=14;O=16;H=1].

The correct option is A 9.38%Molar mass of Mg(NO3)2.6H2O, =256 gm So, % of Mg =24×100256.4=9.38 %Hence, option A is correct.

The total percentage of magnesium present in magnesium nitrate crystals $M g (N O_{3})_{2} .6 H_{2} O$ is:
[ $M g = 24; N = 14; O = 16; H = 1$ ].

The correct option is A 9.38%
Molar mass of $M g (N O_{3})_{2} .6 H_{2} O$ , $= 256 g m$

So, % of $M g$ $= 24 \times \frac{100}{256.4} = 9.38$ %

Hence, option $A$ is correct.