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Chemistry Question on Some basic concepts of chemistry

If 3.01×10203.01 \times 10^{20} molecules are removed from 98 mg98 \, mg of H2SO4H_2SO_4, then number of moles of H2SO4H_2SO_4 left are

A

0.1×10−3 mol0.1 \times 10^{-3} \, mol

B

0.5×10−3 mol0.5 \times 10^{-3} \, mol

C

1.66×10−3 mol 1.66 \times 10^{-3} \, mol

D

9.95×10−3 mol9.95 \times 10^{-3} \, mol

Answer

0.5×10−3 mol0.5 \times 10^{-3} \, mol

Explanation

Solution

The correct answer is B:0.5×10−3 mol0.5 \times 10^{-3} \, mol
98 mg98 \,mg of H2SO4=6.02×1023×10−3=6.02×1020H _{2} SO _{4}=6.02 \times 10^{23} \times 10^{-3}=6.02 \times 10^{20}
Number of H2SO4H _{2} SO _{4} left
=6.02×1020−3.01×1020=6.02 \times 10^{20}-3.01 \times 10^{20}
=1020(6.02−3.01)=3.01×1020=10^{20}(6.02-3.01)=3.01 \times 10^{20}
∵ n=NNA\because \, n =\frac{N}{N_{A}}
=3.01×10206.02×1023=\frac{3.01 \times 10^{20}}{6.02 \times 10^{23}}
=0.5×10−3 mol=0.5 \times 10^{-3}\, mol