10) An empty 149 ml flask weighs 68 g before a sample of volatile liquid is adde

Question

10) An empty 149 ml flask weighs 68 g before a sample of volatile liquid is added. The flask is then placed in a hot (85C) water bath with barometric pressure at 740 torr. The liquid vaporizes and the gas fills the flask. After cooling, flask and condensed liquid together weigh 68.697 g. What is the molar mass of the liquid. 11) A chemical engineer places a mixture of noble gases consisting of 5.50 g of He, 15.0 g of Ne, and 35.0g of Kr in a piston-cylinder assembly at STP. Calculate the partial pressure of each gas. 12) Solid lithium hydroxide is used to scrub CO2 from the air in spacecraft submarines. It reacts with CO2 to produce lithium carbonate and water. What mass of lithium hydroxide is required to remove 215 L of CO2 at 23 C and 0.942 atm 2LiOh(s) +CO2(g)- Li2CO3(s) +H2O(l)

Explanation / Answer

10.

We know that

PV = nRT

P = Pressure = 740 torr
V = Volume = 149 mL = 0.149 L
R = gas constant = 62.364 L Torr mol-1 K-1
T = Temperature = 85 oC = (273 + 85)K = 358 K

So,

PV = nRT

n = PV / RT
   = [ (740 torr) (0.149 L) ] / [ (62.364 L Torr mol-1 K-1) (358 K) ]

   = 0.005 mol

Now,

Mass of the liquid = 68.697 g – 68 g = 0.697 g

So, Molar mass = Mass / moles
                          = 0.697 g /0.005 mol

                          = 139.4 g/mol

11.

So divide by the atomic mass to get molar ratios. And that will determine the partial pressures.

Now, moles are

He = 5.50 / 4 = 1.375 mol
Ne = 15.0 / 20.18 = 0.743 mol
Kr = 35.0 / 83.80 = 0.418 mol

So, Total moles = (1.375 + 0.743 + 0.418) mol = 2.536 mol

Since it's at STP, the total pressure = 1 atm.

So, the partial pressure terms are

PHe = (1.375 / 2.536) atm = 0.542 atm
PNe = (0.743 / 2.536) atm = 0.292 atm
PKr = (0.418 / 2.536) atm = 0.165 atm

12.

2LiOH(s)   +   CO2(g)   ----->    Li2CO3(s)   +   H2O(l)

In the above reaction equation,

2 moles of LiOH reacts with 1 mole of CO2.

Now, for CO2

PV = nRT

P = Pressure = 0.942 atm
V = Volume = 215 L
R = gas constant = 0.082 L atm mol-1 K-1
T = Temperature = 23 oC = (273 + 23) K = 296 K

So,

PV = nRT

n = PV / RT
   = [ (0.942 atm) (215 L) ] / [ (0.082 L atm mol-1 K-1) (296 K) ]
   = 8.34 mol

Now,

Since, 1 mole of CO2 reacts with 2 moles of LiOH
8.34 mole of CO2 reacts with (8.34 x 2) moles of LiOH
8.34 mole of CO2 reacts with 16.68 moles of LiOH

Molar mass of LiOH = 24 g/mol

So, 1 mole of LiOH = 24 g
16.68 moles of LiOH = 16.68 x 24 g = 400.32 g

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