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# What is the density of carbon tetrachloride vapor at STP?

How do I find the density of carbon tetrachloride vapor at STP?

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STP means standard temperature and pressure

T = 273 K

P = 1 atm

We need to find the density = (mass / volume)

We can find the volume of the gas if we use the ideal gas law

(pressure) x (volume) = (# of moles) x (temperature) x R

V = (n x T x R) / (P)

Because density is an intensive property (it does not change with the amount), we will assume one mole of CCl4

So now

P = 1 atm

T = 273 K

n = 1 mole

R = 0.0821 (L*atm) / (Mole *K)

V = (1 mole x 273 K x 0.0821((L*atm)/(mole*K))) / (1 atm) = 22.4 L

This volume is expected with STP of any ideal gas

Now we need mass. We assumed 1 mole of CCl4 to find the volume, so we find the mass of 1 mole of CCl4

1 mole CCl4 x (154 g/mol) = 154 g CCl4

Now we solve for density

D = m/V = 154 g/22.4 L = 6.875 g/L

0.006875 g/ mL

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• Density Of Carbon Tetrachloride

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• Molar mass CCl4 = 153.8236 g/mol

this mass of vapour has volume of 22.4 litres at STP

Mass of 1 litre at STP = 153.8236/22.4 = 6.86g/litre

Answer: Density of CCl4 gas at STP = 6.86g/litre.

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• PV = nRT and n= m / MM combine two eq. and the fact that density = mass/volume, you get. also at stp P=1atm and T= 273.15 K

PV = mRT/MM --> PxMM= d RT --> d= PxMM / RT

d = (1 atm) (molar mass of CCl4) / (0.08206) (273.15)

Source(s): My knowledge
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