Answer:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
![\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}](https://tex.z-dn.net/?f=%20%5Cmathfrak%7B%20%5Cfrac%7B3.8%20%7D%7B8%7D%20%5Ctimes%2010%20%5E%7B24%7D%7D%20%20%20)
<u>= 4.75 × 10²³ molecules of Propane</u>.
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No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
![\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }](https://tex.z-dn.net/?f=%20%5Cmathsf%7B%20%5Cfrac%7B44%20%7D%7B6.02%20%5Ctimes%20%20%7B10%7D%5E%7B23%7D%20%7D%20%5Ctimes%204.75%20%5Ctimes%20%20%7B10%7D%5E%7B23%7D%20%20%7D)
![\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }](https://tex.z-dn.net/?f=%20%5Cmathsf%7B%20%20%3D%20%5Cfrac%7B44%20%7D%7B6.02%20%5Ctimes%20%20%20%5Ccancel%7B%7B10%7D%5E%7B23%7D%20%7D%7D%20%5Ctimes%204.75%20%5Ctimes%20%5Ccancel%7B%20%7B10%7D%5E%7B23%7D%7D%20%20%7D)
![\mathsf{ = \frac{44 }{6.02 } \times 4.75 }](https://tex.z-dn.net/?f=%20%5Cmathsf%7B%20%20%3D%20%5Cfrac%7B44%20%7D%7B6.02%20%7D%20%5Ctimes%204.75%20%20%20%7D)
<u>= 34.72 g</u>