Question

10) How many grams are there in 1.00 x 10^24 molecules of BC13?

Answer 1

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 $\frac{1}{6.022*10^{23} }$ moles.

So $1*10^{24} molecules = \frac{1*10^{24} }{6.022*10^{23} } =1.66 moles$

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

$Molecular mass of BCl_{3}= Mass of Boron + (3*Mass of chlorine)$

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.

$grams of BCl_{3}=Molar mass of BCl_{3}*Number of moles$

$grams of BCl_{3}=117.17*1.66=194.5 g$

So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.