SOLUTIONS

 

 

Problem 1

Using a platform balance, the mass of an irregular block of iron is found to be 280.2 g. The iron was immersed in water and found to displace 35.6 mL of the water. Determine the density of the iron.  

 

The first thing we need to do is to change the 35.6 mL to cm3 since we usually express volume as cubic

centimeters.

 

Since we know than 1 mL = 1cm3,   we can change it

                       

                                             35.6 mL •   

 

1cm3
  mL

 

 

     



     

                                                                       

                                             

 

35.6 mL
      1

 

 

  •   

1cm3
   mL

     

                     

                                             

 

35.6 mL
      1

 

 

  •   

1cm3
   mL

  =   35.6 cm3    

 

 

Now let’s use the formula for Density

D =     

 

m
V

 

 

     



     

 

 

 

 

But remember we have to round to the necessary number of significant figures (sig figs).  In this case, we have to round to three sig figs because of the 35.6 had 3 sig figs.  So our answer is --------------à

D =     

 

m
V

 

 

  =    

 280.2 g
35.6 cm3

  = 7.8707 g/cm3   

 

 

 

7.87 g /cm3

 

 

           

 

Problem 2

Determine the concentration of a solution in grams of salt per gram of solution when the solution is prepared by dissolving 10.2 g of a certain salt in 500 mL of water at 50 °C.

 

The problem asks us to determine the concentration of a salt in a solution.   We know that we are dealing with a ratio:   mass salt/mass solution. 

 

The mass of the solution is the sum of the mass of salt which is dissolved and the mass of water which is used.  The volume of water at 50°C  is known, but we need to know its mass.  Density relates the mass and volume of a material.

 

The first thing we do is to find the density of water at 50°C.  We look that up in the Density of Water Chart

(Click here) and we find out it is 0.98807 g/mL.

 

Now we need to find the mass of the water.  Using our formula: 

 

 

Since we know the density of the water already, we have to change the density formula around so that we can solve for the mass.   We change it to  ------------------------------------------------à

D =     

 

m
V

 

 

     



     

 

 

m = D • V

 

 

      m =   

 

0.98807 g
     mL

 

 

  •   

500 mL water 
        1

     

 

      m =   

 

0.98807 g
     mL

 

 

  •   

500 mL water
        1

  = 494.035 g water   

 

We will round this to 494 g water.  Now we have our mass of water.

 

 

Our mass of salt is given to us already in grams, so we do not have to convert anything.

 

We now have our mass of water= 494 g water and mass of salt = 10.2 g

Remember mass of solutions was the salt and water together.  So let’s add them together and we get 504.2 g

 

Now the problem wants us to determine the concentration of salt in the solution. 

 

So remember our ratio: mass salt/mass solution

 

Using this:  we have                       10.2 g salt                   =  0.0202300 g salt/g solution

                                                          504.2 g solution

But again we have to use our significant numbers to round it off.  We have to round off to 3 sig figs because of the 10.2.  Remembering the rule about zeros, in this case initial zeros—we will not count the zero to the right of the decimal and we will round it 0.0202.  (See Lesson 4 for further explanation about sig figs)

So our answer is 0.0202 g salt/g solution