Question

How many grams are in 5.87x1021 molecules of sulfur?

Answer 1

To find:

Mass of sulphur in 5.87 × 10^21 molecules.

Calculation:

Number of moles of sulphur in the given number of molecules is:

$\therefore \: moles = \dfrac{5.87 \times {10}^{21} }{avagadro \: no.}$

$= > moles = \dfrac{5.87 \times {10}^{21} }{6.023 \times {10}^{23} }$

$= > moles = \dfrac{5.87 \times {10}^{ - 2} }{6.023 }$

$= > moles = 0.97 \times {10}^{ - 2} \: moles$

Now , mass of sulphur in that number of moles:

$\therefore \: moles = \dfrac{given \: mass}{atomic \: mass}$

$= > \: 0.97 \times {10}^{ - 2} = \dfrac{given \: mass}{32}$

$= > \: given \: mass = 32 \times 0.97 \times {10}^{ - 2}$

$= > \: given \: mass = 0.32 \times 0.97$

$= > \: given \: mass = 0.3104 \: grams$

So, final answer is:

$\boxed{ \bf{ \: given \: mass = 0.3104 \: grams}}$