Question

A mixture contains 1.37x10^23 molecules PCl3 and 0.385 mol PCl5- how many grams of chlorine and how many moles of phosphorous are there?

Answer 1

Given the number of molecules of PCl3 = 1.37x10²³  

The mass of Avagadro's number(=6.023x10²³) of molecules of PCl3 = molecular mass of PCl3 = 137.3 g

Hence the mass of 1.37x10²³ molecules of PCl3 = (137.3 g / 6.023x10²³) x 1.37x10²³ = 31.23 g

Hence moles of PCl3 equivalent to  1.37x10²³ PCl3 molecules = mass / mlecular mass

= 31.23 g / 137.3 g/mol = 0.2275 mol PCl3 in the mixture.

The mass of chlorine atom present in 1 mole of PCl3 = 3 x atomic mass of Cl = 3 x 35.45 g = 106.35 g Cl/mol

Hence mass of chlorine atom present in 0.2275 mole of PCl3 = 106.35 g Cl /mol x 0.2275 mol = 24.19 g Cl

Also given moles of PCl5 in the mixture = 0.385 mol

The mass of chlorine atom present in 1 mole of PCl5 = 5 x atomic mass of Cl = 5 x 35.45 g = 177.5 g Cl/mol

Hence mass of chlorine atom present in 0.385 mole of PCl5 = 177.5 g Cl /mol x 0.385 mol = 68.34 g Cl

Hence total mass of Chlorine in the mixture = mass of Cl in PCl3 + mass of Cl in PCl5

= 24.19 g + 68.34g = 92.53 g Cl = 92.5 g Cl (answer)

Moles of P present in 1 mole of PCl3 = 1 mole P

=> Moles of P present in 0.2275 mol PCl3 = 0.2275 mol P

Moles of P present in 1 mole of PCl5 = 1 mole P

=> Moles of P present in 0.385 mol PCl5 = 0.385 mol P

Hence total moles of P = 0.2275 mol + 0.385 mol = 0.612 mol P (answer)