Question

atm 4 Mass of benker and displaced water 368.910 S Mass of empty beaker 144.502 Mass of displaced water (W4 - N5) 224. MRE 7 Barometric pressure 745,mm Hg HO Vapor pressure at 22.0°C 19.8 mm Hg Pressure of H2 (745.0 mm - 19.8 mm) 725.2 mm Hg 10 Pressure of Hy in atmospheres Density of displaced water 0.99823 mL Volume of displaced H2O (mass + density) Volume of H2 - Volume of displaced H2O mL Volume in liters 15 / Ambient temperature 22.0°C 16 Ambient temperature in Kelvin Find the number of moles of H2 using PV = nRT →n-PV/RT (V in liters, Pin atm, T in K, R=0.0821 L-atm/mol-K). mols H2. Find the number of moles of H2 per gram of sample (mols / m). _mols Hyg Find the % Al in the sample using the graph prepared for Problem 1. _%AI Find the % Al in the sample using [%A1/100) x 0.0556] + {[(100-%AI)100] x 0.0153: How work here and on back of this sheet.

Answer 1

We know that the 760mmHg = 1 atm

Pressure in atmosphere = ( Pressure in mmHg / 760 mmHg) atm

=(725.2mmHg / 760mmHg) atm

= 0.9542atm

The density of displaced water = 0.99823g/ml

Mass of displaced water =224.402g

We know; Density = Mass / Volume

The volume of displaced water = mass / density

= 224.402g / 0.99823g ml^-1

= 224.8ml

volume in liters = volume in ml / 1000

= 224.8ml / 1000

= 0.2248L

Temperature in Kelvin = Ambient temperature + 273

=(22.0 + 273) K

= 295K

Using PV = nRT

We have, 0.9542atm * 0.2248L = n *(0.0821 L atm/ mol K )* 295K

n =0.0088moles

Now, Moles per gram = No. of moles of sample present / total mass of the sample present

Moles per gram of hydrogen = 0.0088moles / 224.402g

= 3.9 * 10^-5 mol/g

Hope this helps!