194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have [tex]\frac{1}{6.022*10^{23} }[/tex] moles.
So [tex]1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles[/tex]
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be
[tex]Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)[/tex]
Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.
[tex]grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles[/tex]
[tex]grams of BCl_{3}=117.17*1.66=194.5 g[/tex]
So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
