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The percentage weight of Zn in white vitriol [ZnSO₄ 7H₂O] is approximately equal to Zn=65, S=32, O=16 and H=1)
Options
(a) 33.65%
(b) 32.56%
(c) 23.65%
(d) 22.65%
Correct Answer:
22.65%
Explanation:
Mol. Wt of ZnSO₄ 7H₂O = 65 + 32 + (4 x 16) + 7 (2 x 1 +16)=287
% Mass of Zinc = 65/287 x 100 = 22.65%
Related Questions: - The oxyacid of sulphur that contains a lone pair of electrons on sulphur is
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Question Type: Memory
(964)
Difficulty Level: Easy
(1008)
Topics: Basic Concepts of Chemistry
(94)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The oxyacid of sulphur that contains a lone pair of electrons on sulphur is
- Which is correct about zero order reaction
- The PSLV rocket used
- The isomerism represented by ethyl acetoacetate is
- Aspirin is a acetylated product is
Question Type: Memory (964)
Difficulty Level: Easy (1008)
Topics: Basic Concepts of Chemistry (94)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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