A student obtained the following data in this experiment. Fill in the blanks in the data and make the indicated calculations

Mass of gelatin capsule 0.1284 g Ambient temperature, t 22°C
Mass of capsule plus alloy sample 0.3432 g Ambient temperature, T K
Mass of alloy sample, m 0.2148 g Barometric pressure 745 mm Hg Mass of empty beaker 144.5 g Vapor pressure of H2O at t mm Hg
Mass of beaker plus displaced water 368.9 g Pressure of dry H2, 745 mm Hg Pressure of dry H2 atm
Mass of displaced water 224.4 g Volume of displaced water (density = 1.00 g/mL) mL Volume, V, of H2 = Volume of displaced water mL; liters

Find the number of moles of H2 evolved, (V in liters, P in atm, T in K, R = 0.0821 literatm/ mole K). moles H2
Find , the number of moles of H2 per gram of sample.

Ambient temperature K = 295 K

Pressure of dry H2 (atm) = 0.980 atm

Volume of displaced water (mL) = 224.4 mL

Volume of displaced water in liters = 0.2244 L

Number of mole of H2 = 0.00908 mol

Moles of H2 per gram sample  = 0.042 mol H2/g

Explanation:

Given data:

Mass of gelatin capsule = 0.1284 g;     Ambient temperature (T) = 22°C

Mass of capsule plus alloy sample = 0.3432g; Ambient temperature K = ?

Ambient temperature K = 22+273

                                       = 295 K

Mass of alloy sample = 0.2148;   Barometric pressure 745 mm Hg

Mass of beaker plus displaced water =368.9 g

Pressure of dry H2 = 745 mm Hg

Pressure of dry H2 in atm  = ?

Pressure of dry H2 (atm) = 745 mmHg *1 atm/760 mmHg

                                         = 0.980 atm

Mass of displaced water = 224.4 g

density = 1.00 g/mL)

Volume of displaced water = ?

Density = mass/volume

Volume = mass/density

             =224.4 g/1.  g/mL

           = 224.4 mL

Therefore, volume of displaced water = 224.4 mL

Volume of displaced water in liters = 224.4*10^-3 L

                                                          = 0.2244 L

Calculating the number of mole of H2 using ideal gas law, we have;

PV = nRT

n = PV/TT

where;

n = number of mole of H2

P = pressure of H2 = 0.980 atm

V = volume of H2 = 0.2244 L

R = gas constant = 0.0821 literatm/ mole K

T = temperature =  295 K

Substituting into the formula, we have;

n = PV/RT

  = 0.980* 0.2244/ 0.0821*295

  = 0.219912/24.2195

  = 0.00908 mol

Therefore, number of mole of H2 = 0.00908 mol

Calculating the number of moles of H2 per gram sample using the formula;

Moles of H2 per gram sample = Moles of H2/Mass of sample

                                                  =  0.00908 /0.2148

                                                   = 0.042 mol H2/g

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