탄자니아에서의 화학 수업교안(Mole concept and Related calculation)2

chapter 4. The Mole concept and Related Calculations

→ This page is the 2nd page of chapter 4

Now teach you concentration of Solution. I will tell you that as follow.

◇ A solution is a liquid mixture in which the solute is uniformly distributed within the solvent.

◇ An aqueous solution is a solution of a solute in water.

 

※Molar solutions : A molar solution is a solution which contains one mole of a solute is dissolved in one litre of solution.

 

for example

ex1. A molar solution(1M) of sulphuric acid(H2SO4) contains 98g of H２SO4 in a litre of solution.

 

ex2. A molar solution(1M) of Sodium hydroxide(NaOH) contains 40g of NaOH in a litre of solution.

 

** Note that we are referring to a litre of the solution and not a litre of solvent being added to the solute.

 

NaOH + H2O → NaOH

solute : NaOH

solvent : H2O

solution : aqueous NaOH

 

※ Concentration of solutions

concentration of solution is the amount of a solute in a given volume of solution. The quantity could be in terms of grams or moles.

 

4.6 Molar concentration

Molar concentration (molarity) is the concentration of solution expressed in mole per litre(mol/litre) or mol/dm-3

 

Molarity = Number of moles/ volume of solution(dm3)

Molarity = concentration g dm-3/ molar mass (gmole-3)

Q1. calculate the number of moles of solute in 20 cm3 of 2M sodium hydroxide solution.

solution : 1litre = 1dm3 = 1000cm3

1000cm3 of solution contains 2 moles of hydroxide solution. so 20cm3 of solution contains x moles of sodium hydroxide solution

∴ x= 20 × 2/1000 = 0.04moles

 

∴ 20 cm3 of 2M sodium hydroxide solution contains 0.04 mol of sodium hydroxide.

 

Q2. 25cm3 of a solution contains 0.94g Na2CO3. what is the concentration of the solution in mol dm3? (Na2CO3 1mol=106g)

sol) 25cm3 of the solution contains 0.94g of Na2CO3

25 : 0.94 = 1000 : x

x=37.6g Molarity= 37.6/106 =0.36M

so, Concentration in moles dm-3 of Na2CO3 solution is 0.36M

 

-----Stoichiometric calculations ----

Q3. Calculate number of moles of hydrochlolic acid that reacts with 0.25 moles of calcium carbonate. CaCO3 + 2HCl → CaCl2 + H2O +CO2

 

 

ans) Number of moles of HCl is 0.5M moles.