Question

what volume of CCl4 (d=1.6g/cc) contain 6.02Ã10^25 CCl4 molecules?

Answer 1

Let

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.

Answer 2

Answer: $9625cm^3$

Explanation:

According to Avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.02\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {Avogadro's no}}$

$\text{Number of moles}=\frac{6.02\times 100^{25}}{6.02\times 10^{23}}=100moles$

Now 1 mole of $CCL_4$ molecule weighs = 154 g

100 moles of $CCl_4$ molecule contains = $\frac{154}{1}\times 100=15400g$

Now, $Density=\frac{Mass}{Volume}$

$1.6gcm^{-3}=\frac{15400g}{Volume}$

$Volume=9625cm^3$