Question

How many grams are there in 1.18 x 1024 molecules of boron trihydride, BH3?

Answer 1

6.022×10^23 molecules = 117.17 g

1 molecules = 117.17/ 6.022×10^23

So, 1×10^24 molecules = (117.17×1×10^24)/(6.022×10^23)= 194.569 g