Question

what volume of carbon tetrachloride contains 6.02 ×10^25 carbon tetrachloride molecule(cl= "35.5)"​

Answer 1

Explanation:

Mass of CCl4, m = 154g

w = unknown weight of CCl4 in question

A mole of CCl4 contains 4 × 6.022 × 10^23 chlorine atoms

No of moles × 4 × 6.022 × 10^23 = 10^25

no. of moles, n = w/m = w/154

ie, w = (154 × 10^25) ÷ (4 × 6.022 × 10^23 )

we get, w = 640g (approx)

Let ‘v’ be the unknown volume and ‘d’ be the density

d = w ÷ v

d = 1.5 g/cc

v = 640 ÷ 1.5 = 427 cc (approx)

It would require 427 cc or 0.427 L of CCl4 to contain 10^25 atoms of chlorine.