Which of the following contains same number of carbon atoms as are in 6.0g of carbon (C-12)?

Options

A. 6.0 g ethane
B. 8.0 g methane
C. 21.0 g Propane
D. 28.0 g CO

Answer

B

Explanation

(а) 6.0 g of ethane (`C_(2)H_(6)` )
`C_(2)H_(6)` = molar mass = 24 + 6 = 30g
30 g of ethane contains `2xx6.023xx10^(23)` Carbon atoms.
(b) 8.0 g of methane (`CH_(4)` )
`CH_(4)` = molar mass = 12 + 4 = 16 g 16 g of methane contains `6.023xx 10^(23)` Carbon atoms.
(c) 21.0 g of propane (`C_(3)H_(8)` )
`C_(3)H_(8)` = molar mass = 36 + 8 = 44 g
44 g of propane contains `3 xx6.023xx 10^(23)` Carbon atoms.
(d) 28.0 g of Carbon monoxide (CO)
CO = molar mass = 12 + 16 = 28 g
28 g of Carbon monoxide contains `6.023xx 10^(23)` Carbon atoms.
6.0 g of Carbon contains `(6.023xx10^(23))/12xx 6 = 3.0115xx10^(23)` Carbon atoms.
Among the (a), (b), (c), (d)-8g g of `CH_(4)` contains `(6.022xx10^(23))/16xx8`
`=3.0115xx10^(23)`carbon atoms