Dilution Calculator


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CHEMIX School concentration and dilution calculator allow efficient dilution calculations. A minimal amount of steps is needed for this kind of calculations.
It is possible to use molar, molal, w/w%, w/v%  or mole fraction (x) stock concentrations .

Dilution Calculator


Useful formulas:      C1 V1  = C2 V2               Mnsolute/Vsolution                m = nsolute / msolvent
 

Dilution Calculation Example 1

Knowing that the density of a 17.4M CH3COOH solution = 1.05148 g/mL
a) How many mL of  a 17.4M a acetic acid stock solution is needed for making 3500 mL of a 2M acetic acid solution.
b) How many grams of  a 17.4M a acetic acid stock solution is needed for making 3500 mL of a 2M acetic acid solution.


                    We know that:             M =  n/Vsolution                        C1 V1  = C2 V2      (dilution formula)
Solution a)

17.4M *   C2  =  3.5L  *  2M    

V2 =  (3.5L * 2M)/17.4M = 0.4023 L = 402.3mL

Solution b)

Since we already know  (from solution a) )  the volume of  a 17.4M solution needed and the density we can calculate the mass :

We need to measure out   402.3mL * 1.05148 g/mL= 423.01g  of 17.4M acetic acid


CHEMIX School Solution: a) - dilution by volume  and b) - dilution by mass

1) Select acetic acid (Substance frame)
2) Insert 17.4 and select Molar in the Dilution-Stock concentration frame
3) Insert Vsolu.  = 3500mL  in the Amount frame
4) Insert 2 in  the Molarity (M) text field (Concentration frame)

Procedure - dilution


      ******** Dilution by volume ********
Put about 2000 mL of water into a flask and add 402.299 mL acetic acid stock conc. Add water until 3500 mL
       ******** Dilution by mass ********
Put about 2000 g of water into a flask and add 423.009 g of acetic acid stock conc. Add water until 3553.06 g


  
Dilution Calculation Example 2


Prepare 200mL of a 2 molal solution by diluting a 5 Molar sodium chloride solution.
              FWNaCl=58.4425g/mol  , density of a 2m solution = 1.07417 g/mL ,  density of water = 1g/mL



Solution:

Convert from molal to molar and solve the problem using:    C1 V1 = C2 V2  

Molarity of a 2molal solution (using 1Kg solvent for simplicity):

m= n/msolvent     m = 2mol/1Kg     msolution = 1Kgsolvent+2n  = 1000g+ 2*58.4425g = 1116.885g

Volume of solution:  Vsolution =  1116.885g/1.07417 g/mL = 1039.765mL

M = n/Vsolution      C2 = 2/1.039765Kg  = 1.92351M

V1 = C2V2/C1 = 200mL*1.92354M/5M = 76.94mL

Solution:

Prapare a 2 molal (m) sodium chloride solution by diluting 76.94 mL of the stock conc.(5Molar) solution to 200mL.
 

CHEMIX School Solution:

1) Select sodium chloride (Substance frame)
2) Insert stock concentration = 5 and select Molar in the Dilution-Stock concentration frame
3) Insert Vsolu.  = 200mL  in the Amount frame
4) Insert 2 in  the molality (m) text field (Concentration frame)


Procedure - dilution

      ******** Dilution by volume ********
Put about 90 mL of water into a flask and add 76.9404 mL sodium chloride stock conc. Add water until 200 mL
       ******** Dilution by mass ********
Put about 90 g of water into a flask and add 91.2324 g of sodium chloride stock conc. Add water until 214.834 g



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