Concentrated hydrochloric acid is usually available at a concentration of $ $ 37.7 $ $ percent by mass. The density of the solution is $ 1.19gm{l^{ - 1}} $ . What is its molar concentration?
Answer
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Hint: We know that molarity is referred to as the number of moles (n) of solute present in one litre of the solution. To solve this question, we will first find the volume of the solution using the relation between mass, volume and density of the solution. Then, using the number of moles of solute and volume of the solution, we will find the molar concentration of the solution.
Complete Step By Step Answer:
We know that the molar concentration is referred to as the number of moles (n) of solute present in one litre of the solution. And the formula of the molarity is given as:
$ M = \dfrac{{n \times 1000}}{{V(mL)}}\_\_\_\_\_(i) $
In the question it is given that $ $ 37.7 $ $ percent hydrochloric acid by mass which can be interpreted as $ $ 37.7g $ $ of HCl present in the $ $ 100g $ $ of the solution. Now, to find the molar concentration, we will have to find the volume of the solution using the formula which is given as:
$ Volume = \dfrac{{mass}}{{density}} $
Here, density is given as $ 1.19gm{l^{ - 1}} $ and the mass of the solution is $ $ 100g $ $ . Therefore the volume will be:
$ Volume = \dfrac{{100}}{{1.19}} $
$ Volume = 84.033ml $
Therefore, volume of the solution is found to be $ 84.033ml. $
Now, we have to find the moles of HCl and as we know that number of moles (n) of solute (HCl) can be found by dividing the given mass by the molecular mass of HCl. Now, the mass of Hydrochloric acid is given as $ $ 37.7g $ $ and the molecular weight of HCl is $ $ 36.5g $ $ . So, after putting the values in the formula, we will get:
$ n = \dfrac{{37.7}}{{36.5}} $
$ n = 1.0328 $
Now, we will put the value of moles (n) and the volume of solution (i) in the equation (1), we will get:
$ M = \dfrac{{1.0328 \times 1000}}{{84.033}} $
$ M = 12.29M $
Thus, the molar concentration of the solution will be $ 12.29M. $
Note:
There is one more concentration term which is known as normality and we should not get confused between the terms molarity and normality. Normality is defined as the number of moles of solute which are present in one kilogram of the solvent whereas molarity is referred to as the number of moles (n) of solute present in one litre of the solution.
Complete Step By Step Answer:
We know that the molar concentration is referred to as the number of moles (n) of solute present in one litre of the solution. And the formula of the molarity is given as:
$ M = \dfrac{{n \times 1000}}{{V(mL)}}\_\_\_\_\_(i) $
In the question it is given that $ $ 37.7 $ $ percent hydrochloric acid by mass which can be interpreted as $ $ 37.7g $ $ of HCl present in the $ $ 100g $ $ of the solution. Now, to find the molar concentration, we will have to find the volume of the solution using the formula which is given as:
$ Volume = \dfrac{{mass}}{{density}} $
Here, density is given as $ 1.19gm{l^{ - 1}} $ and the mass of the solution is $ $ 100g $ $ . Therefore the volume will be:
$ Volume = \dfrac{{100}}{{1.19}} $
$ Volume = 84.033ml $
Therefore, volume of the solution is found to be $ 84.033ml. $
Now, we have to find the moles of HCl and as we know that number of moles (n) of solute (HCl) can be found by dividing the given mass by the molecular mass of HCl. Now, the mass of Hydrochloric acid is given as $ $ 37.7g $ $ and the molecular weight of HCl is $ $ 36.5g $ $ . So, after putting the values in the formula, we will get:
$ n = \dfrac{{37.7}}{{36.5}} $
$ n = 1.0328 $
Now, we will put the value of moles (n) and the volume of solution (i) in the equation (1), we will get:
$ M = \dfrac{{1.0328 \times 1000}}{{84.033}} $
$ M = 12.29M $
Thus, the molar concentration of the solution will be $ 12.29M. $
Note:
There is one more concentration term which is known as normality and we should not get confused between the terms molarity and normality. Normality is defined as the number of moles of solute which are present in one kilogram of the solvent whereas molarity is referred to as the number of moles (n) of solute present in one litre of the solution.
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