Question
Question: How much concentrated solution do you measure out and how much water do you need to add to make \( 5...
How much concentrated solution do you measure out and how much water do you need to add to make 500ml 0.1MHCl from 2MHCl ?
Solution
Molar concentration is a measure of the concentration of a chemical substance, in terms of amount of the substance per unit volume of the solution. The SI unit of molarity is moldm - 3 . However, when a solution has a concentration of 1mol/L it is said to be a 1 molar solution and is represented as 1M of the particular chemical substance that forms the solution.
Formulas used: We will be using the formula to find the number of moles of a substance provided we have the molarity of its solution, n=MV where n is the number of moles of the substance, M is the molarity of the substance, and V is the volume of the solution in litres.
Complete Step by Step answer
We know that the molarity is the measure of concentration of a chemical substance in its solution. A substance that has molarity 1M is nothing but that one mole of the chemical substance is dissolved in 1 litre of the solvent to form a solution with concentration 1M . Thus, from the definition of molarity we can infer that,
1M=1L1mole
From the problem we can infer that 0.1M solution of HCl is made with 500ml volume of water, we need to find the amount of HCl that should be added to the 2M concentrated solution. This means that the number of moles of the chemical substance remains unchanged. Thus,
M1V1=n=M2V2
M1V1=M2V2
Substituting the known values of M1=2M,V1=x,M2=0.1M,V2=500mL we get,
(2.0M)×(x)=(0.1)×500mL
Solving for V1=x ,
x=2.00.1×500
⇒x=250=25mL
Thus, to make a 2M concentrated solution we have to dilute 25mL in 475mL of water.
Note
In a laboratory the process can be done using a volumetric flask to ensure the total volume is 500mL . Also make sure that the water is not poured into the acid, this causes an exothermic explosive reaction. Concentration can also be expressed as molality, i.e. the number of moles of solute present in 1Kg of solvent.