
How much water must be added to $ 1500mL $ of a $ 6.0mol/L $ $ CaC{l_2} $ solution to make the concentration of the resulting solution $ 2.0mol/L $ ?
Answer
531.3k+ views
Hint :Basically, concentration is the quantity of solute present in the given quantity of solution and we normally define it in terms of molarity. To give a better idea if you want to make a lemonade if you add more sugar the concentration is more and it is more sweet in taste but as soon as you add water it gets less concentrated that is the concentration decreases.
Complete Step By Step Answer:
Molar quantity of $ 6.0mol/L $ and $ 1500mL $ volume. Now we want to dilute this concentration by $ 2.0mol/L $
We know how to find out concentration,
$ concentration = \dfrac{{moles}}{{volume}} $
As we know in most of the cases the solvent is water sometimes when it comes to gases it is not.
Now we cannot filter out calcium salt out of the mixture but we can dilute it by adding more water to it.
And to find out how much water we need to add, we can use basic relationship,
$ {C_1}{V_1} = {C_2}{V_2} $ Where, $ {C_1} = $ concentration and $ {V_1} = $ volume
Now we know initial and final concentration and we know initial volume also,
So $
{V_2} = \dfrac{{{C_1}{V_1}}}{{{C_2}}} \\
= \dfrac{{6 \times 1500}}{2} = 4500 \\
{V_2} = 4500mL \\
$
Since, we have initial volume $ 1500mL $ we will have to add more $ 3000mL $ to get $ 4500mL $ .
Note :
The overall study of concentration of solutions is a very important part in chemistry. While studying chemical reactions it tells us how much molecules collide at which speed that means it describes the speed of any chemical reaction. Concentration of solutions is defined in terms of molarity, mole fraction, molality, normality, formality and many more.
Complete Step By Step Answer:
Molar quantity of $ 6.0mol/L $ and $ 1500mL $ volume. Now we want to dilute this concentration by $ 2.0mol/L $
We know how to find out concentration,
$ concentration = \dfrac{{moles}}{{volume}} $
As we know in most of the cases the solvent is water sometimes when it comes to gases it is not.
Now we cannot filter out calcium salt out of the mixture but we can dilute it by adding more water to it.
And to find out how much water we need to add, we can use basic relationship,
$ {C_1}{V_1} = {C_2}{V_2} $ Where, $ {C_1} = $ concentration and $ {V_1} = $ volume
Now we know initial and final concentration and we know initial volume also,
So $
{V_2} = \dfrac{{{C_1}{V_1}}}{{{C_2}}} \\
= \dfrac{{6 \times 1500}}{2} = 4500 \\
{V_2} = 4500mL \\
$
Since, we have initial volume $ 1500mL $ we will have to add more $ 3000mL $ to get $ 4500mL $ .
Note :
The overall study of concentration of solutions is a very important part in chemistry. While studying chemical reactions it tells us how much molecules collide at which speed that means it describes the speed of any chemical reaction. Concentration of solutions is defined in terms of molarity, mole fraction, molality, normality, formality and many more.
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