What volume of decimolar solution should be mixed in a solution of same solute to make it 0.25 molar

Definitions

Solute-the substance being dissolved

Solvent-the substance doing the dissolving (the larger amount)

Solution- a homogeneous mixture of the solute and the solvent

Solution= solvent + solute

Aqueous (aq)= water solution

Tincture= alcohol solution

Amalgam= Mercury solution

Molarity (M)- is the molar concentration of a solution measured in moles of solute per liter of solution.

The molarity definition is based on the volume of the solution, NOT the volume of water.

Vocab. Lesson

 Incorrect= The solution is 5.0 Molarity.

Correct= The solution is 5.0 Molar.

 Example Problems

Level 1- Given moles and liters

Determine the molarity when 3.0 moles of sucrose are dissolved to make 2.0 liters of solution.

Level 2- Given Molarity and liters of solution

 Determine the number of moles of salt dissolved in 5.0 liters of a 0.50M solution of salt water.

cross multiply, X= 2.5 mols

Level 3- Given grams (instead of moles) and liters of solution

Determine the molarity when 117g of NaCl are dissolved to make 0.500 liters of solution.

   1st convert to moles, 2nd plug into the molarity equation

117g NaCl( 1mol/58.5g)= 2.00mol NaCl

Level 4-Given grams (instead of moles) and milliliters of solution (instead of liters)

Determine the molarity when 55.5g of CaCl2 are dissolved to make 250.mL of solution.

1st convert to moles, 2nd convert to liters, 3rd plug into the molarity equation

55.5g CaCl2( 1mol/111g)= 0.500mol CaCl2

250.ml( 1L/1000mL) =0.250L

Past Regents Questions-Follow link to check the answers

Jan 2003-44 What is the molarity of a solution of NaOH if 2 liters of the solution contains 4 moles of NaOH?

(1) 0.5 M     (3) 8 M
(2) 2 M       (4) 80 M

Jan. 04-41 What is the molarity of a solution containing 20 grams of NaOH in 500 milliliters of solution?

(1) 1 M (2) 2 M (3) 0.04 M (4) 0.5 M

Jan 2002-42 What is the molarity of a solution that contains 0.50 mole of NaOH in 0.50 liter of solution?

(1) 1.0 M      (3) 0.25 M
(2) 2.0 M      (4) 0.50 M

Aug. 2006-42 How many total moles of KNO3 must be dissolved in water to make 1.5 liters of a 2.0 M solution?

     (1) 0.50 mol     (2) 2.0 mol     (3) 3.0 mol     (4) 1.3 mol

Aug 2005-

41 What is the total number of moles of NaCl(s) needed to make 3.0 liters of a 2.0 M NaCl solution?(1) 1.0 mol      (3) 6.0 mol

(2) 0.70 mol    (4) 8.0 mol

June 2006-

16 Molarity is defined as the(1) moles of solute per kilogram of solvent(2) moles of solute per liter of solution(3) mass of a solution

(4) volume of a solvent

Aug 2008-

15 Which phrase describes the molarity of a solution?(1) liters of solute per mole of solution(2) liters of solution per mole of solution(3) moles of solute per liter of solution

(4) moles of solution per liter of solution

June 2009-46 Which sample of HCl(aq) contains the greatest number of moles of solute particles?
(1) 1.0 L of 2.0 M HCl(aq)
(2) 2.0 L of 2.0 M HCl(aq)
(3) 3.0 L of 0.50 M HCl(aq)
(4) 4.0 L of 0.50 M HCl(aq)

June 2007-

13 A 3.0 M HCl(aq) solution contains a total of(1) 3.0 grams of HCl per liter of water(2) 3.0 grams of HCl per mole of solution(3) 3.0 moles of HCl per liter of solution

(4) 3.0 moles of HCl per mole of water

June 2010-14 The molarity of an aqueous solution of NaCl is defined as the(1) grams of NaCl per liter of water(2) grams of NaCl per liter of solution(3) moles of NaCl per liter of water

(4) moles of NaCl per liter of solution

Jan 2008-

15 Which unit can be used to express solution concentration?(1) J/mol     (3) mol/L

(2) L/mol    (4) mol/s

Jan 04-41 What is the Molarity of a solution containing 20 grams of NaOH in 500 milliliters of solution?(1) 1 M      (3) 0.04 M

(2) 2 M      (4) 0.5 M

Jan 2010-40 What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF, (gram-formula mass =26 grams/mole)?(1) 1.3 M    (3) 3.0 M

(2) 2.0 M    (4) 0.75 M

on to ppm or Molality

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Text Solution

Answer : 50 mL

Solution : The number of gram moles in 50 mL `M/5` HCI. <br> `=(1/5 xx 50)/1000 =0.01` (No. of gram moles `=(M xx V)/1000`) <br> Suppose, v mL of water should be added to make it decimolar. <br> The total volume will now be = 50 + v mL. <br> Number of gram moles of HCI in this diluted solution <br> `=(1/10 xx (50 + v))/1000` <br> The two gram moles should be equal. Therefore, <br> `=(1/10 xx (50 + v))/1000 = 0.01` <br> which gives v=50 mL <br> Hence, 50 mL of water should be added to 50 mL `M/5` HCI to make it decimolar.

The most important thing to keep in mind when it comes to diluting solutions is that the number of moles of solute must remain constant at all times.

Simply put, the number of moles of solute present in the dilute solution must be equal to the number of moles of solute present in the concentrated sample.

This is the key to any dilution calculation. As you know, molarity is defined as

#color(blue)("molarity" = "moles of solute"/"liters of solution")#

In essence, when you dilute a solution, you decrease its concentration by increasing its volume, which in turn is done by adding more solvent to the solution.

#color(white)(a)#

#color(white)(a)#

So, you know the molarity and volume of the target solution, which means that you can determine how many moles of hydrochloric acid it must contain

#color(blue)(c = n/V implies n = c * V)#

#n_(HCl) = "0.040 mol" color(red)(cancel(color(black)("L"^(-1)))) * 1.00color(red)(cancel(color(black)("L"))) = "0.040 moles HCl"#

Now your task is to determine what volume of the concentrated solution would contain this many moles of hydrochloric acid.

#color(blue)(c = n/V implies V = n/c)#

Plug in your values to get

#V = (0.040color(red)(cancel(color(black)("moles"))))/(0.25color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.16 L"#

Expressed in milliliters and rounded to two sig figs, the answer will be

#V = color(green)("160 mL")#

This is exactly what the formula for dilution calculations allows you to do

#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution"))#

Here you have

#c_1#, #V_1# - the molarity and volume of the concentrated solution
#c_2#, #V_2# - the molarity and volume of the diluted solution

Once again you will have

#V_1 = c_2/c_1 * V_2#

#V_1 = (0.040color(red)(cancel(color(black)("M"))))/(0.025color(red)(cancel(color(black)("M")))) * "1.00 L" = color(green)("160 mL")#

So, you would prepare this solution by adding enough water to your #"160-mL"# sample of #"0.25-M"# hydrochloric acid solution to make the total volume of the resulting solution equal to #"1.00 L"#.