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

The Percentage Weight Of Zn In White Vitriol Znso 7ho Chemistry Question

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:

  1. Among the following the paramagnetic one is
  2. Which of the following is heavy water
  3. The volume of 2.8g of carbon monoxide at 27⁰C and 0.821 atm pressure is
  4. Polyamide
  5. Which is the correct order of increasing energy of the listed orbitals in the atom

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.

NEETLab Mobile App

Share this page with your friends

Be the first to comment

Leave a Reply

Your email address will not be published.


*