Review of Solution Preparation Techniques in Chemistry.
You want to mix a 30% alcohol solution with a 70% alcohol solution to make 400mL of 60% alcohol solution. How much 70% solution should you use?
METHOD 1 :
70% of some quantity ( Solution x ) out of one container + 30% of some quantity ( Solution y ) out of the other = ( 60% )( 400mL ). Let’s put this into a mathematical format.
First, ( 70%)( x mL ) + ( 30% )( y mL ) = ( 60% )( 400mL ). Since 60% and 400mL are known quantities, we may begin by writing this equation as ( 70% )( x mL ) + ( 30% )( y mL ) = 240mL ( alcohol ). Note that 60% = ( 60 mL alcohol / 100 mL solution ) x 100% before converting the percentage to its appropriate decimal value.
Now, let’s deal with the remaining percentage values ( 70% and 30% ) that need to be converted to decimal format. 70% is equivalent to ( 70mL alcohol / 100mL solution ) x 100%. Likewise, 30% is equivalent to ( 30mL alcohol / 100mL solution ) x 100%.
Let’s now convert 70% alcohol by volume and 30% alcohol by volume to their appropriate decimal form. These decimals will effectively work as conversion factors regarding how much solution is needed to get the final quantity of alcohol that is desired. ( NOTE: The 100% noted above is not necessary to derive an appropriate decimal expression ).
( 70 mL / 100mL )( x mL ) + ( 30mL / 100mL )( y mL ) = 240mL.
Appropriate substitution into Equation 1 yields ( 0.7 )( x mL ) + ( 0.3 )( y mL ) = 240mL.
Now, if some percentage of ( x mL ) + some percentage of ( y mL ) = 240 mL, the unknown quantity ” y ” can be rewritten in terms of ” x ” to simplify things. If x + y = 240mL, then 240mL – x = y. We can now substitute ( 240mL – x ) in place of y in the above derivation.
( 0.7 )( x mL ) + ( 0.3 )( 400mL – ( x mL ) ) = 240mL.
The remaining steps involve doing algebra without taking any important conceptual considerations into account. After multiplication, we get ( 0.7 )( x mL ) + ( 0.3 )( 400mL ) – ( 0.3 )( x mL ) = 240mL. Next, we have ( 0.7 )( x mL ) + 120mL – ( 0.3 )( x mL ) = 240mL
Let’s now isolate like terms on opposite sides. Expressions with the unknown variable ” x ” will appear on the left, and the known values will appear on the right hand side ( RHS ) of the equation. ( 0.7x mL ) – ( 0.3x mL ) = 240mL – 120mL. Further simplification leads to 0.4x mL = 120mL
x mL = 300mL
We must take 300 mL out of the bottle of 70% alcohol.
METHOD 2 :
We want to know how much alcohol ( unknown quantity = x ) is taken from a 70% solution and mixed with an unknown quantity ( y ) taken from a 30% solution in order to obtain a 60% by volume alcohol solution in a 400mL container.
Let’s attempt to do this in as abstract a manner as possible. 70% of solution ” x ” plus 30% of solution ” y ” = 60% of solution ” z “. Put into the appropriate mathematical formula, the above circumstance is denoted as ( 70% )( x ) + ( 30% )( y ) = ( 60% )( z ).
Recall that parentheses denote multiplication between things in parentheses. Mathematical operations that occur within parentheses ( none here ) would occur before multiplication between terms in parentheses occur. Let’s attempt to get an answer without converting the percentages to decimals. We can isolate the x by dividing all three terms by 70%. This will leave us with the following equation : x + ( 3 / 7 )( y ) = ( 6 / 7 )( z ).
According to conservation theories regarding any conservative principle ( mass, charge, energy, momentum, etc. ), if exact proportions of x, y, and z are needed to make things work out, a relationship between x, y, and z can be justified with the statement that x + y = z. If this is the case, then z – x = y. Thus, the value for y above can be replaced with z – x.
Therefore, x + ( 3 / 7 )( z – x ) = ( 6 / 7 )( z ). Let’s now multiply to expand the middle term. x + ( 3 / 7 )( z ) – ( 3 / 7 )( x ) = ( 6 / 7 )( z ).
Things can be made simpler by using algebra to get terms containing ” z ” on the right-hand side ( RHS ) of the equation and leaving terms containing ” x ” on the left-hand side ( LHS ) of the equation.
We now have x – ( 3 / 7 )( x ) = ( 6 / 7 )( z ) – ( 3 / 7 )( z ).
Note that ANYTHING DIVIDED BY ITSELF = 1. This justifies placing a factor of ( 7 / 7 ) in front of the ” x ” term above. This will make subtraction of ( 3 /7 )( x ) from ” x ” much simpler.
Dividing every term by 7 yields ( 7 / 7 )( x ) – ( 3 / 7 )( x ) = ( 6 / 7 )( z ) – ( 3 / 7 )( z ).
Further simplification after subtraction gives us ( 4 / 7 )( x ) = ( 3 / 7 )( z )
Dividing both sides by ( 4 / 7 ) yields the following result :
x = ( 3 / 7 )( 7 / 4 )( z )
x = ( 3 / 4 )( z )
Since the original value of ” z ” was given ( 400 mL ), x = ( 3 / 4 )( 400mL ) = ( 1200mL / 4 ) = 300mL
x = 300mL is the volume of alcohol needed to be taken from the 70% solution.
RENAISSANCE PHYSICS 