a) Calculate the volume in cm3 at r.t.p. of 0.3 moles of neon
b) What is the mass of 0.015 moles of sodium sulfate, Na 2 SO 4
c) Calculate the concentration in mol/dm 3 of 0.5 moles of hydrochloric acid in 500
cm 3 of solution
Answers
a) As neon (Ne) is a monoatomic gas.
1 mole of any gas at STP has a volume equal to 22.4 litres.
So, at STP 0.3 moles of Neon will have a volume = 0.3 x 22.4 => 6.72 liters.
As we know, STP conditions are 273 K and 1 atm.
RTP conditions are 25°C (25 + 273 =298 K) and 1 atm.
So, we have to convert 6.72 liters at STP to RTP.
P1 = 1 atm P2 = 1 atm
V1 = 6.72 L V2 = ?
T1 = 273 K T2 = 298 K
By gas equation:
P1V1/T1 = P2V2/T2
1 x 6.72/273 = 1 x V2/298
V2 = 6.72 x 298/273
V2 = 7.335 litres
Hence, at RTP the volume will be: 7.335 litres or 7335 mL .
b) Molar mass of Na2SO4 = 2 x 23 + 32 + 4 x 16 =>46 + 32 + 64
=> 142 grams per mole.
Hence, 0.015 moles will have a mass of 0.015 x 142 => 2.13 grams.
Hence, the mass will be 2.13 grams.
c)
No. of moles = 0.5 moles of HCl
Volume of solution = 500 cm3
1 dm3 = 10^-2 x 1 m3
1 dm3 = 10^-2 x 1000 L
1 dm3 = 10^(-2+3) L
1 dm3 = 10 L
1 dm3 = 1000 cm3
1/2 dm3 = 1000/2 cm3
0.5 dm3 = 500 cm3
Volume of solution = 0.5 dm3
Hence, concentration = 0.5/0.5
=> 1 mol/dm3