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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: - Which of the following is a strong reducing agent
- In Kjeldahl’s mathod, the nitrogen present is estimated as
- In the titration of strong acid and weak base, the indicator used is
- In an isobaric process,when temperature changes from T₁ to T₂, ΔS is equal to
- Sodium Hydroxide reacts with Zinc to form
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
- Which of the following is a strong reducing agent
- In Kjeldahl’s mathod, the nitrogen present is estimated as
- In the titration of strong acid and weak base, the indicator used is
- In an isobaric process,when temperature changes from T₁ to T₂, ΔS is equal to
- Sodium Hydroxide reacts with Zinc to form
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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