Question

What is the percentage concentration by mass of a $ {\text{16 mol }}{{\text{L}}^{ - 1}} $ solution of nitric acid, for which density is $ {\text{1}}{\text{.42 gm m}}{{\text{l}}^{ - 1}} $ ?

Answer 1

Hint: We are given with the molarity of nitric acid as $ {\text{16 M}} $ because the units of molarity of substance is $ {\text{mol }}{{\text{L}}^{ - 1}} $ . Now we will convert the unit of density which is given in $ {\text{gm m}}{{\text{l}}^{ - 1}} $ to $ {\text{gm }}{{\text{L}}^{ - 1}} $ . Then with the help of the unitary method we will find the mass of nitric acid present in the $ {\text{16 mol }}{{\text{L}}^{ - 1}} $ solution.

Complete Step By Step Answer:
Since we know that the molarity of a solution is the ratio of the number of moles of solute and the volume of solution in litres. Its unit will be $ {\text{mol }}{{\text{L}}^{ - 1}} $ . Therefore we can say that we are given the molarity of nitric acid as $ {\text{16 M}} $ . Now we are given the density of solution as $ {\text{1}}{\text{.42 gm m}}{{\text{l}}^{ - 1}} $ . We will convert the units of density from $ {\text{gm m}}{{\text{l}}^{ - 1}} $ to $ {\text{gm }}{{\text{L}}^{ - 1}} $ . We will use the conversion factor as:
 $ \Rightarrow {\text{ 1 L = 1000 mL}} $
Thus we can write as:
 $ \Rightarrow {\text{ 1 mL = 1}}{{\text{0}}^{ - 3}}{\text{ L}} $
Also the given density can be written as:
 $ \Rightarrow {\text{ 1}}{\text{.42 gm m}}{{\text{l}}^{ - 1}} $
 $ \Rightarrow {\text{ 1}}{\text{.42 }}\dfrac{{{\text{gm}}}}{{ml}} $
We can use the above conversion factor here as:
 $ \Rightarrow {\text{ 1}}{\text{.42 }}\dfrac{{{\text{gm}}}}{{{{10}^{ - 3}}{\text{ L}}}} $
 $ \Rightarrow {\text{ 1}}{\text{.42 }} \times {\text{ 1}}{{\text{0}}^3}{\text{ }}\dfrac{{{\text{gm}}}}{{{\text{ L}}}} $
 $ \Rightarrow {\text{ 1420 }}\dfrac{{{\text{gm}}}}{{{\text{ L}}}} $ which can also be written as $ {\text{1420 gm }}{{\text{L}}^{ - 1}} $ .
Hence we can observe that when we take one litre of solution then the mass of nitric acid will be $ {\text{1420 gm}} $ .
Now we will calculate the weight of nitric acid from the molarity equation. When we take one litre of solution then $ {\text{16 mol }}{{\text{L}}^{ - 1}} $ of nitric acid will be equal to $ {\text{16 mole}} $ . Since we now that:
 $ \Rightarrow {\text{ Number of Moles = }}\dfrac{{{\text{mass of solute}}}}{{{\text{Molecular Mass}}}} $
The molecular mass of nitric acid $ \left( {HN{O_3}} \right){\text{ = 1 + 14 + 3}} \times {\text{16 = 63 gm}} $ .
Thus the mass of nitric acid will be calculated as:
  $ \Rightarrow {\text{ mass of solute = Number of Moles }} \times {\text{ Molecular Mass}} $
 $ \Rightarrow {\text{ mass of solute = }}\left( {{\text{16 }} \times {\text{ 63}}} \right){\text{ gm}} $
 $ \Rightarrow {\text{ mass of solute = 1008 gm}} $
Now we can find the percentage of weight by weight from above calculated weight or mass as:
 $ \Rightarrow {\text{ % }}concentration{\text{ }}by{\text{ }}mass{\text{ = }}\dfrac{{{\text{1008 gm}}}}{{1420{\text{ gm}}}} \times 100 $
 $ \Rightarrow {\text{ % }}concentration{\text{ }}by{\text{ }}mass{\text{ = 70}}{\text{.98 % }} $
 $ \Rightarrow {\text{ % }}concentration{\text{ }}by{\text{ }}mass{\text{ = 70}}{\text{.98 % }} \simeq {\text{ 71 % }} $
Thus the percentage concentration by mass is $ 71{\text{ % }} $ .

Note:
The molecular mass of nitric acid is calculated by adding the atomic mass of each element multiplied with its atomicity. The final value of percentage composition is obtained after rounding off the calculated percentage. The method which we use is known as the unitary method by assuming one litre of solution.