Some Basic Concepts Of Chemistry Question 31

Question: If $ 3.01\times 10^{20} $ molecules are removed from 98 mg of $ H _2SO _4 $ then the number of moles of $ H _2SO _4 $ left are

Options:

A) $ 0.1\times {10^{-3}} $

B) $ 0.5\times {10^{-3}} $

C) $ 1.66\times {10^{-3}} $

D) $ 9.95\times {10^{-2}} $

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Answer:

Correct Answer: B

Solution:

[b] Moles of $ H _2SO _4 $ in 98 mg of $ H _2SO _4 $ $ =\frac{1}{98}\times 0.098=0.001 $ Moles of $ H _2SO _4 $ removes $ =\frac{3.01\times 10^{20}}{6.02\times 10^{23}}=0.5\times {10^{-3}}=0.0005 $ Moles of $ H _2SO _4 $ left $ =0.001-0.0005=0.5\times {10^{-3}} $



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