Question

what volume of CCl4 (d=1.6g/cc) contain 6.02×10^25 CCl4 molecules?

Answer 1

Answer:

427cc

Explanation:

Volume of CCl4 (d=1.6g/cc)

Let mass of CCl4, m = 154g

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

Thus, the number of moles × 4 × 6.022 × 10^23 = 10^25

Number of moles, n = w/m = w/154

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

= w = 640g (approx)

Let the unknown volume be = v and and the density be = d

d = w/v

d = 1.5 g/cc

v = 640/1.5

v = 427 cc

Therefore, It will require around 427 cc or 0.427 L of CCl4 to contain 10^25 atoms of chlorine molecules

Answer 2

$6.022 \times 10^{23} \text { molecules }=1 \text { moles }$

$6.022 \times 10^{25} \text { molecules }=\left(1 \text { mole } / 6.022 \times 10^{23} \text { molecules) } \times 6.022 \times 10^{25} \text { molecules }\right$  

Hence $6.022 \times 10^{25} \text { molecules }=100\ \text {moles}$

$1 \text { mole } \mathrm{CCl}_{4}=154\ \mathrm{g}$

$100 \text { moles } \mathrm{CCl}_{4}=154 \times 100=15400\ \mathrm{g}$

$=\text { Mass / density }=15400 \text { g } / 1.6 \mathrm{g} / \mathrm{cc}=9,625 \mathrm{cc}$

Hence 9625 cc volume of $\mathrm{CCl}_{4}(\mathrm{d}=1.6 \mathrm{g} / \mathrm{cc}) \text { contain } 6.02 \times 10^{25}\ \mathrm{CCl}_{4}\ molecules$.