| ⇦ | 
| ⇨ |
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: - Tollen’s test can be used to distinguish
- Predict the product of reaction of I₂ with H₂O₂ in basic medium
- Which of the following is wrongly matched
- Phenyl isocyanide test is used to identify
- When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g) each at S.T.P,
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
- Tollen’s test can be used to distinguish
- Predict the product of reaction of I₂ with H₂O₂ in basic medium
- Which of the following is wrongly matched
- Phenyl isocyanide test is used to identify
- When 22.4 litres of H₂(g) is mixed with 11.2 litres of Cl₂(g) each at S.T.P,
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
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Leave a Reply