Question

Usually when we discuss solutions, we talk about their concentration. The most common concentration unit is Molarity. However, there are a few other different measurements of concentration. Please explain what defines molarity, and then also explain what these other concentration units are and how they are defined.

If I started with 1.5L of 6.0M HCl solution, and I diluted it out to 15L – what is the concentration of my diluted HCl? If this diluted solution then was neutralized by 30. mL of KOH – how strong is the KOH solution? (It will be helpful to write out the balanced reaction between HCl and KOH.)

Answer 1

Molarity(c₂) ; It is the amount (mol) of the solute present in a unit volume of the solution .If n₂ mol of of solute are present in V dm³ of a solution then

c₂ = n₂/V mol.dm^-3

here n₂ = w₂/M₂ =(mass of solute )/(molar mass of the solute )

when volume of the solution is expressed as cm³ ,then molarity may be calculated by using the formula

c₂  =(n₂*1000)/V mol.dm^-3

There are a few other different measurements of concentration are

1. NORMALITY

2.MOLALITY

3.MASS FRACTION

4.VOLUME FRACTIONS

5.MOLE FRACTION

NORMALITY (N); The normality of a solution is the number of equivalent of solute per unit volume of the solution .if w₂ is the mass of the solute ,and k is the number of protons ,electron or their equivalents that are lost or gained by solute in the particular reaction of interest ,then

N₂ = (k*w₂)/(M₂*V) =kc

here V is the volume in dm^{3 ,} M₂ is the molar mass of the solute ,W₂  is the mass of the solute .The nomality is always an integral muitiple of the molarity

For HCl = H⁺ + Cl^- , k is one

For H₂SO₄   = 2H⁺ + SO₄^2- k is 2

MOLALITY (m); The molality is the mol of solute per kilogram of solvent .If w₂ is the mass of solute dissolved in w₁ mass unit of solvent then

m₂ =w₂/(M₂ *w₁₎  = n₂/w₁ mol kg ^-1

if the mass of solvent is expressed in grams then molality is given by

m₂   = ( n_{2 *}1000)/w₁ mol.kg^-1

MOLE FRACTION(Xi) ; It is defined as the number of moles of a component present in unit mole of solution .Let w₁  w₂ w_{3 ....} etc ,be the mass of various components 1 ,2 , 3 .....in a solution and M₁ M₂ M₃ .... etc are their respective molar masses then the mole fraction of i_th component is expressed as ,

X_i = n_i /(n₁+ n₂+ n₃ ......)

MASS FRACTION ;It is the mass of a component per unit mass of the solution .if W₁ W₂ are the masses of the two components then the mass fraction w is given by

w = w₁ /(w₁+w₂) or w_{2 /}(w₁+w₂)

VOLUME FRACTION(x);It is the volume of a pure component in a unit volume of the solution .thus if V₁ V₂  V₃ etc. are the volumes of pure individual components ,then the fraction of component 2 is expressed as

x =v₂ / (v₁ +v₂ +v₃)

QUS ;2

If you started with 1.5L of 6.0M HCl solution, and I diluted it out to 15L – what is the concentration of my diluted HCl

let ,initial volume of HCl =v₁=1.5L

initial strength of HCl solution =s₁=6.0 M

final volume =v₂=15 L  

final concentration of Hcl solution =s=?

we know s₁v₁=s₂v₂ that is , 1.5*6 =15 *s2 or s₂=(1.5*6)/15   or  s_{2=0.6 M}

the concentration of my diluted HCl  is 0.6 M

QUS 3;

If this diluted solution then was neutralized by 30. mL of KOH – how strong is the KOH solution

HCl +KOH =KCl+H₂O

Concentration of hcl solution (s₁)=0.6 M

volume of HCl soiution (v₁)=15 ml

volume of KOH solution (v₂)=30 ml

concentration of KOH solution(s₂)=?

now applying equivalence equation

s_1*v₁=s_2*v_{2 or}  0.6*15=30 * s_{2 or} s₂=(0.6*15)/30 _or   s₂₌0.3 M