Question

When x molecules are removed from 200 mg of
N20, 2.89 * 10-³ moles of N20 are left. x will be
10²⁰ molecules
(2) 10¹⁰ molecules
(3) 21 molecules
(4) 10²¹ molecules
​

Answer 1

Answer:

4)1021 molecules

Here, initial moles of N2O=44200×10−3=4.55×10−3

moles removed = initial - final =4.55×10−3−2.89×10−3 

=1.66×10−3moles

∴ no of molecules removed =6.022×1023×1.66×10−3 

=9.996×1020 ≈10×1020

≈1021molecules

Hence, the correct option is (4)