Question #50761

2 Answers
Dec 14, 2017

#30.072 L#

Explanation:

We know that,

The atomic mass of #He# is #4 g=1 mol#

At #STP,#

the molar volume of every gasses of #1mol# is #22.4L#

Then,at #STP#

The volume of #4 g He # is#22.4L#

#:.#The volume of #5.37 gHe # is
#=(5.37 gxx22,4L)/(4 g)=30.072 L#

Dec 14, 2017

See the simple steps to take in solving the question above;

Explanation:

We are dealing on two given information..

Volume and Mass..

Using this we can get the required Volume..

Recall;

#n = v/(mV) = m/(mM)#

Where;

#n = "no of moles"#

#v = "volume"#

#m = "mass"#

#mV = "molar volume"#

#mM = "molar mass"#

So we have;

#v/(mV) = m/(mM)#

#v = ?#

#mV = 22.4dm^3mol^-1or Lmol^-1#

N/B #rArr dm^3 = L#

#m = 5.37g#

#mM = 4gmol^-1#

Substitute the values into the fomula

#v/(22.4dm^3mol^-1) = (5.37g)/(4gmol^-1)#

#v/(22.4dm^3cancel(mol^-1)) = (5.37cancelg)/(4cancelgcancel(mol^-1))#

#v/(22.4dm^3) = (5.37)/(4)#

Cross multiplying..

#v xx 4 = 22.4dm^3 xx 5.37#

#4v = 22.4dm^3 xx 5.37#

#4v = 120.288dm^3#

Divide both sides by #4#

#(cancel4v)/cancel4 = (120.288dm^3)/4#

#v = (120.288dm^3)/4#

#v = 30.072dm^3 or 30.072L#