How much water is to be added to 75ml of alcohol so that the mixture contains 25% of alcohol?

Answer

Hint: In this question, the concept of percentages will be used. The relation between the percentage and the amount of substance in a mixture is given by-
$$\dfrac{\mathrm{Volume}\;\mathrm{of}\;\mathrm{dissolved}\;\mathrm{substance}\;}{\mathrm{Total}\;\mathrm{volume}\;\mathrm{of}\;\mathrm{liquid}}\times100=\mathrm{percentage}$$Complete step-by-step solution -
Now, we have been given that initially 15% of alcohol is present in a 400 mL solution. We will first find the volume of alcohol in the initial solution using the formula. Let the amount of alcohol be v mL, then-
$\dfrac{\mathrm v}{400}\times100=15\\$
v = 60 mL
This is the initial amount of alcohol. Now, let us assume that x mL of alcohol is added to the solution. The new volume of alcohol will be 60+x and the new volume of solution will be 400+x. It is given that the strength of this solution is 32%. So, the equation formed will be-
$\dfrac{60+\mathrm x}{400+\mathrm x}\times100=32$
$\Rightarrow 60 + x = 0.32(400 + x)$
$\Rightarrow 60 + x = 128 + 0.32x $
$\Rightarrow 0.68x = 68 $
$\Rightarrow x = 100 mL $
Therefore, 100 mL of alcohol should be added to make the strength 32%. This is the required answer.

Note: In this question, no specific formula is needed, we just need to apply some logic and express the percentage in terms of volume. Also, writing the right units in the answer is a must.

Key Questions

• The percentage concentration of any solution is most commonly expressed as mass percent:

  Mass % of any component of the solution =
  (Mass of the component in the solution / Total mass of the solution) x 100

  Other methods are:

  1. Volume percentage:

  Volume % of a component =
  (Volume of the component/Total volume of the solution) x 100

  1. Mass by volume percentage:
     It is the mass of solute dissolved in 100 mL of the solution.

  i.e. Mass by Volume percentage =
  (Mass of solute in grams/Volume of solution in mL) x 100

  Here's a point to be kept in mind :
  Whenever we say mass or volume of the solution, you need to add the respective masses and volumes of ALL the components of the solution. Do NOT commit the error of taking the mass or volume of only the solute or solvent in the denominators of the above expressions.

  The concentration of a solution is most of the time expressed as the number of moles of solute present in 1 Liter of the solution (also called molarity )

  (There are also other ways to express concentration. Please follow this link. )

  EXAMPLE:
  (a) If 25 moles of NaCl are present in 100 L of a solution wherein H2O is the solvent, then the concentration of the solution is #25/100=0.25 "mol·L"^-1#.

  (b) What is the molarity of a solution prepared by dissolving 15.0 g of sodium hydroxide in enough water to make a total of 225 mL of solution?

  Solution

  • Calculate the number of moles of solute present.

  Moles of NaOH = 15.0 g NaOH × #(1"mol NaOH")/(40.00"g NaOH")# = 0.375 mol NaOH

  • Calculate the number of litres of solution present.

  Volume = 225 mL × #(1"L")/(1000"mL")# = 0.225 L soln

  • Divide the number of moles of solute by the number of litres of solution.

  Molarity = #(0.375"mol")/(0.225"L")# = 1.67 mol/L

• Let's address the question for both percent concentration by mass and for percent concentration by volume.

  Percent concentration by mass is defined as the mass of solute divided by the total mass of the solution and multiplied by 100%. So,

  #c% = m_(solute)/(m_(solution)) * 100%#, where

  #m_(solution) = m_(solvent) + m_(solute)#

  There are two ways to change a solution's concentration by mass

  • Adding more solute - making the solution more concentrated;
  • Adding more solvent - making the solution more dilute;

  Let's take an example to better illustrate this concept. Say we dissolve 10.0g of a substance in 100.0g of water. Our concentration by mass will be

  #c% = (10.0g)/(10.0g + 100.0g) * 100% = 9.09%#

  Now let's try doubling the mass of the solute; the new concentration will be

  #c% = (2 * 10.0g)/(2*10.0g + 100.0g) * 100% = 16.7%#

  However, if we keep the mass of the solute at 10.0g and doubled the mass of the solvent (in this case, water), the concentration will be

  #c% = (10.0g)/(10.0g + 2*100.0g) * 100% = 4.76%#

  The same is true for percent concentration by volume, which is defined as the volume of the solute divided by the total volume of the solution and multiplied by 100%.

  #c_(volume)% = V_(solute)/(V_(solute) + V_(solvent)) * 100%#

  It's easy to see that manipulating either the volume of the solute or the volume of the solvent (or both) would change the solution's percent concentration by volume.

• There are two types of percent concentration: percent by mass and percent by volume.

  PERCENT BY MASS

  Percent by mass (m/m) is the mass of solute divided by the total mass of the solution, multiplied by 100 %.

  Percent by mass = #"mass of solute"/"total mass of solution"# × 100 %

  Example

  What is the percent by mass of a solution that contains 26.5 g of glucose in 500 g of solution?

  Solution

  Percent by mass =

  #"mass of glucose"/"total mass of solution" × 100 % = (26.5"g")/(500"g")# × 100 % = 5.30 %

  PERCENT BY VOLUME

  Percent by volume (v/v) is the volume of solute divided by the total volume of the solution, multiplied by 100 %.

  Percent by volume = #"volume of solute"/"total volume of solution"# × 100 %

  Example

  How would you prepare 250 mL of 70 % (v/v) of rubbing alcohol

  Solution

  70 % = #"volume of rubbing alcohol"/"total volume of solution" × 100 %# × 100 %

  So

  Volume of rubbing alcohol = volume of solution × #"70 %"/"100 %"# = 250 mL × #70/100#

  = 175 mL

  You would add enough water to 175 mL of rubbing alcohol to make a total of 250 mL of solution.

How much water is added to 75 ml of alcohol so that the mixture contains 25 of alcohol?

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