Question

354.5 g of chlorine gas (MW = 70.9 g/mol) is held in a vessel with a fixed volume of 70. L.

Answer 1

Answer:

1.77 atm

Explanation:

We have to check the values that gives the problem:

V= 70 L

mass =354.5 g

Molas weight= 70.9 g/mol

T=30 ºC

P= ?

We can find the moles of chlorine if we use the molar weight:

$354.5g~\frac{1~mol}{70.9~g}$

$5~mol$

Now, we have the moles, volume, temperature therefore we can use the ideal gas equation:

$P*V=n*R*T$

We know the R value:

$0.082\frac{atm*L}{mol*K}$

We have “K” units for the temperature, so we need to do the conversion:

$30+273.15=303.15~K$

With all the data we can plug the values into the equation:

$P*70L=5mol*303.15K*0.082\frac{atm*L}{mol*K}$

$P=\frac{5mol*303.15K*0.082\frac{atm*L}{mol*K}}{70L}$

$P=1.77~atm$

I hope it helps!