Determination of Molar Volume at S.T.P

Date of experiment: 07/04/2007

Background

Magnesium is an active metal that readily reacts with hydrochloric acid to produce hydrogen gas. In this experiment, we will react a known amount of magnesium with hydrochloric acid and collect and measure the volume of hydrogen gas produced. From the volume of the gas measured at atmospheric pressure and the temperature of the lab, we can calculate the universal ideal gas constant.

Procedure

1- Measure a length of 4 cm of magnesium ribbon, weigh it and record is mass.

2- Connect the ribbon to the mouth of a sealed burette by a 15cm copper wire.

3- Add around 6-7 cm3 of concentrated HCl solution into the burette (keep the ribbon away from the HC1), then fill it with distilled water up to the brim.

4- Lower the ribbon into the water, cover the mouth of the burette with your finger, and invert it into a 400 cm3 beaker two-thirds full of water. Clamp the burette into position on the ring stand.

5- When the reaction comes to an end cover the mouth of the burette with your finger and move it into a 1000 cm3 graduated cylinder full of water. Equalize the levels of the liquids inside and outside the burette. Read the volume of the collected gas, and then measure the temperature of the water in the cylinder.

Apparatus:

Data collection

Quantitative data:

Upper reading of burette: Vi= 50.00cm3 ±0.02

Lower reading of burette: Vf =14.90cm3 ±0.02

Temperature: T1= 20.7 ºC ±0.01

Volume of water at tip of burette (closed side of burette) that has no reading i.e. volume of water in space above reading that indicates 50.00cm3: Vw=1.28±0.02 cm3

Pressure of atmosphere: Patm=754 ±1mmHg

Qualitative data:

While reaction was occurring, the ribbon of magnesium became very shiny.

We could see the HCl slowly diffusing down the burette until it reached the magnesium ribbon. We could track the HCl as it moved down the burette since it formed a type of mist in water.

Data processing and presentation

Determination of volume of H2 gas liberated:

Upper reading of burette: Vi= 50.00 ±0.02 cm3

Lower reading of burette: Vf =14.90 ±0.02 cm3

Volume of water at tip of burette: Vw= 1.28 ±0.02 cm3

Volume of H2 gas liberated:

VH2= Vi -Vf + Vw = (50.00 ±0.02 cm3)-(14.90 ±0.02 cm3)+ (1.28±0.02cm3)

=36.38±0.06cm3

Determination of pressure exerted by H2 gas:

From the adjacent diagram (2) we can say that since the water level at points A and B are the same, the pressure at each of these points is equal to the second.

PA=PB

Pwv +P H2 = Patm

Where Pwv is the pressure exerted by the water vapor, P H2 is the pressure exerted by gas and Patm is the atmospheric pressure

Patm = 754±1mmHg

Pwv = 17.91±0.01mmHg

We need to find the pressure of H2 at point A:

PH2= Patm - Pwv = 736±1mmHg

In Pascal unit: 760.0 mm Hg = 101325 Pa

PH2= 736 ±1mmHg

PH2= (736±1mmHg • 101325 Pa)/ 760mmHg

= 98100±100Pa

Determination of number of moles of H2 liberated:

Mg(s) + 2HCl (aq) --> MgCl2 (aq) + H2 (g)

According to the ideal gas equation of state:

PV=nRT

In these conditions:

Pressure of H2: PH2= 98100±100Pa

Volume: VH2= 36.38cm3 ±0.02=36.38x10^-3 ±0.06x10^-6 m3

Temperature: T1= 20.7 ºC ±0.01

In Kelvin unit: T1=273.15+ (20.7±0.01) = 293.85±0.01 K

R= 8.31 Pa• m3• mol-1• K-1

n (H2)=PH2V/RT1

= (98100±100Pa• 36.38x10^-6m3 ±0.06x10^-6)/ (8.31 Pa• m3• mol-1• K-1• 293.85 K ±0.01)

=0.0015±0.0002mol

Determination of molar volume:

At STP conditions:

P=1.01x10^5 Pa

n= 1 mol

T=273.15 K

(PstpVm)/ (PH2 VH2) = (nRTstp)/ (n (H2) RT1)

Vm = (n stp• R• PH2• VH2• Tstp)/ (nH2• R• Pstp • T1)

= (8.31 Pa• m3• mol-1• K-1• 98100±100Pa• 36.38x10^-6 ±0.06x10^-6 m3• 273.15 K)/ (

0.0014±0.0002mol• 8.31Pa• m3• mol-1• K-1• 1.01x10^5 Pa• 293.85±0.01 K)

=0.0234±0.0004 m3/mol

=23.4±0.4dm3/ mol