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1. Given: mass = 0.334g of C2H6, CAtomicWeight = 12.01, HAtomicWeight = 1.008

Required:

a. Molar Mass of Ethane

b. Number of molecules of C2H6 in 0.334g of C2H6

Formula:

a. (n = (2 x CAtomicWeight) + (6 x HAtomicWeight)

b. mass x (1 mole / a ) x (6.022x1023 molecules / 1 mole)

Solution:

a. Molar Mass of Ethane

n = (2 x 12.01) + (6 x 1.008)

n = 24.02 + 6.0498

n = 30.0698

b. Number of molecules

n = 0.334 x (1 mole / 30.0698 g) x (6.022x1023 molecules / 1 mole)

n = 6.689x1021 molecules

FINAL Answer: The are 6.689x1021 molecules in 0.334g of Ethane

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