Question

how many
atoms
phosphorous (P) Present in 1.4 moles of
pcl3
(a) 4
(b ) 5.6
(e) 8.431 x 1023
Cd) 3.372 X
1024​

Answer 1

Answer:

3.372×1024

Explanation:

1 m

olecule of pcl3 contains 4atoms

Answer 2

Answer:

Formula mass of PCl3 = 31 + 35.5 × 3

= 31 + 106.5

= 137.5 u

1 mole = 6.022×10^23 atoms

1.4 moles = 6.022×10^23 × 1.4 = 8.4308 × 10^23 atoms

since 1 mole of P is present in 1.4 moles of PCl3

no of atoms is 8.4308×10^23 atoms.

hence opt is C

Thank you....

Hope it helps.