Question

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Q. Calculate the difference in the number of carbon atom in 1.4 g of C-14 isotope and 1.0 g of C-12 isotope. ✏️✏️
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Answer 1

Explanation:

No.Of molesin C14 = 1.4/14 =1/10 moles

• no of molecules =
• $6.022 \times 10 to the power 23$

No of moles in C 12 = 1/ 12

no of molecules=

$6.022 \times 10tothe power23$

Difference = 6.022 *10to the power 23(1/10 -1/12)

= 1/60 * 6.022* 10 to the power 23

= 0. 1003 * 10 tothe power 23

= 1.003* 10 to the power 22

Answer 2

Q. Calculate the difference in the number of carbon atom in 1.4 g of C-14 isotope and 1.0 g of C-12 isotope.

$ans = 7.2 \times {10}^{21} carbon \: atoms$

Explanation:-

$14 \: g \: of \: c - 14 \: isotope \: contain \: c \: atoms \: = 6.022 \times {10}^{23}$

$1 \: g \: of \: c - 14 \: contains \: c \: atom \: = \frac{6.022 \times {10}^{23} }{12} \times 1$

$4.30 \times {10}^{22}$

$no. \: of \: atoms \: in \: 1 \: g \: of \: c - 12$

$12 \: g \: of \: c - 12 \: contains \: \: c \: atoms\: = {6.022}^{23}$

$1 \: g \: of \: c - 12 \: contains \: c \: atoms \: = \frac{6.022 \times {10}^{23} }{12} \times 1$

$= 5.02 \times {10}^{22}$

$difference \: in \: no. \: of \: c \: atoms \:$

$=( 5.02 \times {10}^{23}) - (4.30 \times {10}^{23} )$

$= 7.2 \times {10}^{21}$