6. At 99.0°C and 99.7 kPa, a sample of volatile liquid is vaporized completely i

Question

6. At 99.0°C and 99.7 kPa, a sample of volatile liquid is vaporized completely in a 25f6 ml flask. The empty container had a mass of 156.325g. When placed on the balance containing the condensed liquid, its mass was 157 422g. (a) Find the molar mass of the liquid. (b) what is the molecular formula of the liquid it contains 18.0% carbon, 2.26% hydrogen, and 79.7% chlorine? 7. Consider the following reaction: Zn + ta Zna2 (eg) + H2 (g) In the following problem (#8), what is the pressure of the DRY hydrogen gas? 8. Hydrogen is produced by the reaction of zinc with hydrochloric acid. (see above, not balanced). If a sample of hydrogen is collected over water at 758 torr and 21.0°C, how many grams of hydrogen are in a 2.00 L sample of collected, wet hydrogen gas? It takes 22 seconds for methane, CH4, sample to effuse down a capillary tube. Another gas takes 44 seconds. One student thinks that the gas is helium; another student believes it to be sulfur dioxide. Which student is correct? Give mathematical proof 9. 10. A 20.0L container is filled with nitrogen gas at t STP. Then 71.0L of hydrogen gas (measured at STP) is introduced into the container. A spark is set off and the container is cooled to 0.00°C. If the reaction goes to completion (ammonia, NH3 is formed), what will the final pressure in the container be? 11. What volume of sulfur dioxide at 343 C and 1.21 atm is produced by the combustion of 1.83 kg of sulfur?

Explanation / Answer

Volume of flask= 256ml= 256/1000L= 0.256 L

This is also the volume of gas generated, V= 0.256L

T= 99 deg.c= 99+273=372K, P=99.7Kpa= 99.7/101.3 atm=0.984 atm

From gas law, n= PV/RT=0.98*0.256/(0.0821*372)=0.008 moles

Mass of condensed vapor= mass of flask+liquid- mass of flask

= 157.422-156.325 gm =1.097 gm

Moles =mass/molar mass

Molar mass = mass/moles = 1.097/0.008=137.5 g/mole

Atomic weights : C= 12, H= 1and Cl=35.5

Mass % : C:H:Cl= 18: 2.26: 79.7

Mole = mass/moles

Mole ratio = 18/12 : 2.26/1 :79.7/35.5

= 1.5: 2.26: 2.24

Dividing by 1.5 the equation becomes

1: 2,25/1.5: 2.24/1.5= 1: 1.5: 1.%

2: 3:3

Empirical formula is C2H3Cl3

Molar mass of empirical formula = 24+3+3*35.5=137.5

So molar mass of empirical formula =molar mass of compound.

Hence molecular formula is C3H3Cl3.

2.

2. Vapor pressure of water at 21 deg.c= 18.65 mm Hg

Partial pressure of hydrogen = total pressure- vapor pressure of water at 21 deg.c

= 758-18.65 =739.35 mm Hg= 739.35/760 atm =0.973 atm

partial pressure is defined as the pressrue exerted by the gas if it alone occupes all the volume.

P= 0.973 atm, V= 2L, T= 21+273= 294K

From gas law, PV= nRT, R = gas law constant =0.0821 L.atm/mole.K

No of moles, from gas law , n= PV/RT =0.973*2/(0.0821*294)= 0.081

Mass of Hydrogen = moles of hydrogen* molar mass =0.081*2=0.162 gm

3.

Rate of effusion (R) is inversely proportional to sqrt of molar mass

For two different gases having R1 and R2 as rate of effusion and M1, M2 as molar masses. Molar mass of methane, M1=4

R1/R2= Sqrt(M2/M1)

22/44= Sqrt(16/M1)

Taking square ¼= 16/M1

M1= 64g/mole, molar mass of SO2= 64 g/mole, Hence the gas is SO2.

4. 1 mole of any gas occupies 22.4 L at STP

20 L of N2 correspond to 20/22.4 moles =0.893 moles, mole of H2= 71/22.4 =3.2 moles

The reaction between N2 and H2 is N2+3H2------>2NH3, theoretical molar rati of N2:H2= 1:3

actual molar ratio of N2:H2= 0.893: 3.2= 1:3.6

so excess is H2. So all the N2 and 0.893*3=2.679 moles of H2 react to give 2*0.893=1.786 moles of NH3 and 3.2-2.679= 0.521 mole of H2. total moles= 1.786+0.521=2.31mole of gas

T= 0 deg.c= 0+273= 273K, volume of mixture at   STP= 20+71= 91L

Hence P= nRT/V = 2.31*0.0821* 273/91 =0.57 atm

5. Mass of Sulfur =1.5 kg, moles of sulfur= mass of sulfur/atomic weight= 1.83/32 kg moles =0.0572 kg moles

the reaction of Sulfur reacting with O2 is S+O2----->SO2

1 mole of S give   1 mole of SO2

0.0572 kg moles of S gives 0.057 kg moles of SO2. =0.0572*1000 =57.2 gmoles

n=57.2 gmoles, P=1.21 atm, t=343 deg.c =343+273=616K, R= 0.0821 L.atm/mole.K

V= nRT/P= 57.2*0.0821* 616/1.21= 2391 L

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