If `3.01 xx 10^(28)` molecules are removed from 98 mg of `H_2SO_4,` then number of moles of `H_2SO_4` left are

Options

A. `0.1 xx 10^(-2)` mol
B. `0.5 xx 10^(-3)` mol
C. `1.66 xx 10^(-3)` mol
D. `9.95 xx 10^(-2)` mol

Answer

B

Explanation

98 mg of
`H_2SO_4 = 6.02 xx 10^(23) xx 10^(-3) = 6.02 xx 10^(20)`
Number of `H_2SO_4`left
`= 6.02 xx 10^(29)- 3.01 xx 10^(20)`
`= 10^(20)(6.02 - 3.01) = 3.01 xx 10^(20)`
`because n = (N)/(N_A) = (3.01 xx 10^(20))/(6.02 xx 10^(23)) =0.5 xx 10^(-3)` mol