A 25.0 gram sample of a hydrate of iron(III) phosphate was heated until no more water was driven off. The mass of the anhydrous salt is 16.9 grams. What is the % water in the hydrate?

32.4%

Explanation:

Given parameters:

Mass of hydrated sample  = 25g

Mass of anhydrous salt  = 16.9g

Unknown:

Percentage of water in the hydrate  = ?

Solution:

Obviously, there is a mass difference between the original hydrated sample and anhydrous salt which should indicate clearly that the mass loss is due to the evaporated water from the salt.

   Mass of salt of water = mass of hydrated salt  - mass of anhydrous salt

                                        =  25g  - 16.9g

                                         = 8.1g

Percentage of water  =

                                    =

                                     = 32.4%

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.