Question

IF 10^21 molecules are removed from 200 mg CO2 then the number of moles of CO2 left will be

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Answer 1

Answer:

0.00288 moles (approximately)

Explanation:

All the calculations are approximately correct .

Answer 2

$\sf1 \: mole \: of \: Co_{2} \: ⟶ \: 44 \: g \: of \: Co _{2}$

$\sf \: 6 \times {10}^{23} \: molecule \: of \: Co_{2} \: ⟶ \: 44 g \: Co _{2}$

$\sf \: {10}^{21} \: molecule \: of \: Co_{2} \: ⟶ \:?$

$\displaystyle \qquad \sf \: = \: \frac{{10}^{21} \times 44}{6 \times {10}^{23} }$

$\displaystyle \qquad \sf \: = \: 7.33 \times {10}^{ - 2} g$

$\displaystyle \qquad \sf \: = \: 7.33 \times {10}^{ - 2} \times {10}^{3} g$

$\displaystyle \qquad \sf \: = \: 7.33 \times {10}g$

$\displaystyle \qquad \sf \: = \: 73.3mg$

$\sf Amount \: of \: Co_{2} \: left = 200-73.3$

$\sf \qquad \: = \: 126.7mg$

$\displaystyle \sf No.of \: gram \: molecules = \frac{Wt \: of \: molecule}{Gram \: molecular \: wt}$

$\displaystyle \sf \qquad \: = \: \frac{126.7 \times {10}^{ - 3} \cancel g}{44 \cancel g}$

$\displaystyle \sf \qquad \: = \: 2.8 \times {10}^{ - 3} mol$

$\displaystyle \sf \qquad \: = \: 2.8 \: m. mol$

$\displaystyle \sf \qquad \: = \: 0.0028mg \: \: \: \: \: (1g = 1000mg)$

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