Question

1) 362.11 g of O2 are contained in a 3.00 L high pressure tank at a temperature of 186°C. What is the pressure of the gas in torr?

2) A gas has a given volume. The moles of gas are doubled then the pressure is 1/3 of the original pressure. After the pressure change, the Kelvin temperature quadruples (increases 4 times of the original Kelvin temperature). By how much has the volume changed? Be specific (i.e. the volume is 1/3 of the original volume, the volume is 8 time greater, etc.). If you need more information to answer, explain what information is needed.

3) A helium balloon has a volume
of 2.54 L at 1.04 atm and 24^{o}C. The balloon is then released. Calculate the
volume of the balloon (in L) when the pressure is 312 torr and the
temperature is–47^{o}C.

4) A vapor occupies a volume of 237.2 mL at a temperature of 90.6°C and a pressure of 734.6 torr. The mass of the condensed vapor is 0.978 g. Calculate the molar mass of the vapor (hint: Use the ideal gas law to find the number of moles).

5) A chemist mixed 10.00 g of a sodium carbonate solution with 10.00 grams of a hydrochloric acid solution in an open beaker. The reaction mixture fizzed. The final mass of the mixture was 19.31 g. Did this reaction violate the Law of Conservation of Mass? Why or why not? Be sure to explain your answer for full credit (hint – think common sense).

Answer #1

**1)**

**No of mol of O2 = 362.11/32 = 11.31 mol**

**P = nRT/V = (11.31*0.0821*(273.15+186))/3 =142.11
atm**

** = 1.08*10^5 torr**

**2) n2 = 2 n1**

**p2 = p1/3**

**T2 = 4T1**

**V2 = ?**

**P1V1/T1n1 = p2v2/T2n2**

**let p1 = 1 , n1 = 1 , t1 = 1**

**(1*1/1*1) = (1/3)*V2 / (4*2)**

**V2 = 24V1**

**that means 24 times higher than initial**

**3)**

**2.54*1.04/(297.15) =
(312/760)*V2/(273.15-47)**

**V2 = 4.89 L**

**4) PV = nRT**

**(734.6/760)*(237.2/1000) =
(0.978/x)*0.0821*(273.15+90.6)**

**molarmass of vapour = 127.4 g/mol**

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