Question

@ Q Search prelab continued: A student obtained the following data in this experiment. Perform the calculations needed to fill in the blanks. Show all the steps to your calculations (including correct sig figs and units) in the space below for credit. mass of gelatin capsule mass of capsule + metal 0.1987 g 0.4076 g mass of metal 143.05 g mass of empty beaker mass of beaker + displaced water mass of displaced water 357.86 g 22°C temperature density of water volume of displaced water volume of H2 mLL 759.4 torr Atmospheric pressure vapor pressure of water pressure of dry H2 torr atm moles hydrogen gas evolved moles hydrogen ion used equivalent mass of metal Show the steps to your calculations completely and neatly:

This is a pre-lab and there are no more information given. Only those given number in the boxes.

Answer 1

First 7 questions have been answered here. For the last two parts , we need to know the reaction done in the experiment or identity of the metal.

Concept: In the experiment, volume of water displaced is equal to the volume occupied by H2 gas released. The gas evolved (Hydrogen) is assumed to behave ideally, so PV= nRT holds.

For determining pressure of H2, Dalton's law of partial pressure is used, which means:

Ptotal = Pwater vapor + PH2

Calculations: (Note that final answers are shown with 4 significant digits)

1. mass of metal = (mass of capsule + metal) - mass of capsule

                           = 0.4076 g - 0.1987 g

                         = 0.2089 g

2. mass of displaced water = (mass of beaker + displaced water)- mass of empty beaker

                                             = 357.86 g - 143.05 g

                                            = 214.8 g

3. Density of water at 22 oC = 0.9978 g /cm3         

Note that you can also use the general value of 1 g/cm3. But for exact values, we are using the above value.

4. Volume of displaced water = mass of displaced water / density of water

    (From 2. and 3. above)    = ( 214.81 g) / 0.9978 g /cm3        

= 215.3 cm3      = 215.3 ml

Note : 1 cm3  = 1 ml

5. Since, volume of water displaced is equal to the volume occupied by H2 gas released

Volume of H2 gas =   215.3 ml = 0.2153 L

6. Ptotal = 759.4 torr = Pwater vapor + PH2

At 22 oC , Pwater vapor = 19.8 torr;

So, 759.4 torr = 19.80 torr + PH2

Or,         PH2 = 739.6 torr = 0.9733 atm

As, 1 torr = 0.001316 atm

7. Using; PV= nRT

Or,        n = (PV)/ RT          ;

given T = (22 + 273 ) = 295 K ; P and V are known from parts 6. and 5. respectively.

and using R = 0.0821 L atm mol-1K-1

n = (0.9733 atm) x (0.2153 L) / (0.0821 L atm mol-1K-1 x 295 K )

n   = 0.008652