The percentage weight of Zn in white vitriol -ZnSO4- 7H2O- is approximately equal to-

The correct option is D 22.65% Molecular weight of white vitriol ZnSO_4.7H_2O = 65 + 32 + (4 × 16) + 7(2 × 1 + 16) =287 gWeight of Zn in white vitrio ...

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Standard XII

Chemistry

Introduction to Aldehydes and Ketones

The percentag...

Question

The percentage weight of Zn in white vitriol $[Z n S O_{4} \cdot 7 H_{2} O]$ is approximately equal to?

33.65 %

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32.56 %

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23.65 %

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22.65 %

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Solution

The correct option is D $22.65 %$
Molecular weight of white vitriol $Z n S O_{4} .7 H_{2} O = 65 + 32 + (4 \times 16) + 7 (2 \times 1 + 16) = 287$ g
Weight of Zn in white vitriol = 65 g
Percentage weight of Zn = $\frac{W e i g h t o f Z n}{M o l e c u l a r w e i g h t o f w h i t e v i t r i o l} \times 100 = \frac{65}{287} \times 100 = 22.65$ %

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Similar questions

F e S O_{4} ∙ 7 H_{2} O

(Green vitriol) represents which type of crystal system?

Z n S O_{4} \cdot 7 H_{2} O

loses

H_{2} O

on exposure to air due to efflorescence property.

If true enter 1, else enter 0.

Q. Calculate the percentage of water of crystallisation in green vitriol.

Q. In the reaction of

Z n + F e S O_{4} \to Z n S O_{4} + F e

Q. The chemical formula of white vitriol is:

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The percentage weight of Zn in white vitriol [ZnSO4⋅7H2O] is approximately equal to?

The correct option is D 22.65%Molecular weight of white vitriol ZnSO4.7H2O=65+32+(4×16)+7(2×1+16)=287 gWeight of Zn in white vitriol = 65 gPercentage weight of Zn = WeightofZnMolecularweightofwhitevitriol×100=65287×100=22.65 %

The percentage weight of Zn in white vitriol $[Z n S O_{4} \cdot 7 H_{2} O]$ is approximately equal to?