10g of CO on burning in air gives the product which have the number of nucleons

Options

A. `9.47 xx 10 ^(23)`
B. `94.7 xx 10 ^(23)`
C. `947 xx 10 ^(23)`
D. `947 xx 10 ^(24)`

Answer

B

Explanation

`CO + (1)/(2) O _(2) to CO _(2)`
`{:("1 mole", "1 mole"),(28g, 44g):}`
`10 g CO = (44)/(28) xx 10 g CO _(2) = ( 44 xx 10 )/(28 xx 44) ` mole `CO _(2)`
`= ( 44 xx 10)/( 28 xx 44) xx 6 . 02 10 ^(23)` molecules of `CO _(2) = ( 44 xx 10 xx 6. 02 xx 10 ^(23))/( 28 xx 44) xx 44` nucleons
`= 94. 647 xx 10 ^(23) ` nucleons (Totla n nd p which are present in the nucleus are called nucleons. )