Question

18.066x10²¹ molecules of ethane occupies volune at NTP snd weihghs __gms

Answer 1

Answer:

18.066 × $10^{21}$ molecules of ethane occupy volume at NTP and weigh 0.9 grams.

Explanation:

The molecular mass of ethane = 30 g/mol

That is, one mole of Ethane is equivalent to 30 g. One mole of ethane contains an Avogadro number of particles.

The mass of 6.022 × $10^{23}$ ethane molecules  = 30 g

So the mass of 1 molecule of ethane = $\frac{30}{6.022*10^{23} } = 4.981*10^{-23} g$

Therefore the mass of 18.066 × $10^{21}$ molecules                                                       = $4.981*10^{-23} *(18.066 * 10^{21)} = 0.899 g$

approximately equal to 0.9 grams