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