Answer
Verified
450.3k+ views
Hint: Normality (N) is defined as the number of gram equivalents of a solute present per litre of the solution. It is given as:
\[\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]
if normality of one solution is given, that of other can be calculated using normality equation which is given as
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]
Specific gravity is the ratio of the density of the given substance to the density of water. density of water is 1$gm{{L}^{-1}}$.
Complete answer:
(a) To find the normality of a 96% solution of sulphuric acid, ${{H}_{2}}S{{O}_{4}}$.
- A 96% solution means that 96 grams of ${{H}_{2}}S{{O}_{4}}$ is dissolved in 100 grams of solution.
Given weight of ${{H}_{2}}S{{O}_{4}}$ = 96 grams
Molar mass of ${{H}_{2}}S{{O}_{4}}$ = 98 grams.
Basicity of ${{H}_{2}}S{{O}_{4}}$ is 2. Therefore, equivalent weight will be equal to
\[\begin{align}
& \text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{Basicity}}=\dfrac{98g}{2}=49g\,e{{q}^{-1}} \\
& \\
\end{align}\]
Given specific gravity of ${{H}_{2}}S{{O}_{4}}$= 1.84. specific gravity is given as
\[\begin{align}
& \text{Specific gravity = }\dfrac{\text{density of }{{\text{H}}_{2}}S{{O}_{4}}}{\text{density of }{{\text{H}}_{2}}O} \\
& \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=\text{Specific gravity }\times \text{density of }{{\text{H}}_{2}}O \\
& \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=1.84\times 1=1.84g\,m{{L}^{-1}} \\
\end{align}\]
Density of ${{H}_{2}}S{{O}_{4}}$ in the 100 gram solution =$\dfrac{100}{VmL}$
Volume, V of the solution will be = $\dfrac{100g}{1.84gm{{L}^{-1}}}=54.35mL$
Hence, the normality of 96% ${{H}_{2}}S{{O}_{4}}$ with specific gravity 1.84 is calculated to be
\[\begin{align}
& \text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}} \\
& N=\dfrac{96g}{49g\,e{{q}^{-1}}}\times \dfrac{1000}{54.35mL}=36.048N \\
\end{align}\]
(b) - Normality of 96% ${{H}_{2}}S{{O}_{4}}$, ${{N}_{1}}$= 36.048
Let the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$ required to make 0.1N (${{N}_{2}}$) of solution in 1 litre (${{V}_{2}}$) be ‘${{V}_{1}}$’.
Using normality equation, we have
\[\begin{align}
& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
& {{V}_{1}}=\dfrac{{{N}_{2}}{{V}_{2}}}{{{N}_{1}}} \\
& {{V}_{1}}=\dfrac{0.1N\times 1000mL}{36.05N}=2.77mL \\
\end{align}\]
Therefore, the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$(36.05 N) required to make 0.1N solution in 1 litre (1000 mL) is 2.77 mL.
(c) Using normality equation, we can calculate the volume (${{V}_{2}}$) to which 10 mL (${{V}_{1}}$) 96% solution of ${{H}_{2}}S{{O}_{4}}$(${{N}_{1}}=36.05N$) is diluted to prepare solution of 2 N (${{N}_{2}}$).
\[\begin{align}
& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
& {{V}_{2}}=\dfrac{{{N}_{1}}{{V}_{1}}}{{{N}_{2}}} \\
& {{V}_{1}}=\dfrac{36.05N\times 10mL}{2N}=180.25mL \\
\end{align}\]
Therefore, 10 mL of 36.05N should be diluted to 180.25 mL to make 2N solution.
Note: Convert all the units for volume in either litre or millilitre and use the same unit throughout the calculation to avoid errors. Carefully follow the steps to calculate the volume of the solution from the specific gravity given.
\[\text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}}\]
if normality of one solution is given, that of other can be calculated using normality equation which is given as
\[{{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}}\]
Specific gravity is the ratio of the density of the given substance to the density of water. density of water is 1$gm{{L}^{-1}}$.
Complete answer:
(a) To find the normality of a 96% solution of sulphuric acid, ${{H}_{2}}S{{O}_{4}}$.
- A 96% solution means that 96 grams of ${{H}_{2}}S{{O}_{4}}$ is dissolved in 100 grams of solution.
Given weight of ${{H}_{2}}S{{O}_{4}}$ = 96 grams
Molar mass of ${{H}_{2}}S{{O}_{4}}$ = 98 grams.
Basicity of ${{H}_{2}}S{{O}_{4}}$ is 2. Therefore, equivalent weight will be equal to
\[\begin{align}
& \text{Equivalent weight = }\dfrac{\text{Molar mass}}{\text{Basicity}}=\dfrac{98g}{2}=49g\,e{{q}^{-1}} \\
& \\
\end{align}\]
Given specific gravity of ${{H}_{2}}S{{O}_{4}}$= 1.84. specific gravity is given as
\[\begin{align}
& \text{Specific gravity = }\dfrac{\text{density of }{{\text{H}}_{2}}S{{O}_{4}}}{\text{density of }{{\text{H}}_{2}}O} \\
& \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=\text{Specific gravity }\times \text{density of }{{\text{H}}_{2}}O \\
& \text{density of }{{\text{H}}_{2}}S{{O}_{4}}=1.84\times 1=1.84g\,m{{L}^{-1}} \\
\end{align}\]
Density of ${{H}_{2}}S{{O}_{4}}$ in the 100 gram solution =$\dfrac{100}{VmL}$
Volume, V of the solution will be = $\dfrac{100g}{1.84gm{{L}^{-1}}}=54.35mL$
Hence, the normality of 96% ${{H}_{2}}S{{O}_{4}}$ with specific gravity 1.84 is calculated to be
\[\begin{align}
& \text{Normality}=\dfrac{\text{weight of solute (in grams)}}{\text{equivalent weight}}\times \dfrac{1000}{\text{volume of solution (in mL)}} \\
& N=\dfrac{96g}{49g\,e{{q}^{-1}}}\times \dfrac{1000}{54.35mL}=36.048N \\
\end{align}\]
(b) - Normality of 96% ${{H}_{2}}S{{O}_{4}}$, ${{N}_{1}}$= 36.048
Let the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$ required to make 0.1N (${{N}_{2}}$) of solution in 1 litre (${{V}_{2}}$) be ‘${{V}_{1}}$’.
Using normality equation, we have
\[\begin{align}
& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
& {{V}_{1}}=\dfrac{{{N}_{2}}{{V}_{2}}}{{{N}_{1}}} \\
& {{V}_{1}}=\dfrac{0.1N\times 1000mL}{36.05N}=2.77mL \\
\end{align}\]
Therefore, the volume of 96% solution of ${{H}_{2}}S{{O}_{4}}$(36.05 N) required to make 0.1N solution in 1 litre (1000 mL) is 2.77 mL.
(c) Using normality equation, we can calculate the volume (${{V}_{2}}$) to which 10 mL (${{V}_{1}}$) 96% solution of ${{H}_{2}}S{{O}_{4}}$(${{N}_{1}}=36.05N$) is diluted to prepare solution of 2 N (${{N}_{2}}$).
\[\begin{align}
& {{N}_{1}}{{V}_{1}}={{N}_{2}}{{V}_{2}} \\
& {{V}_{2}}=\dfrac{{{N}_{1}}{{V}_{1}}}{{{N}_{2}}} \\
& {{V}_{1}}=\dfrac{36.05N\times 10mL}{2N}=180.25mL \\
\end{align}\]
Therefore, 10 mL of 36.05N should be diluted to 180.25 mL to make 2N solution.
Note: Convert all the units for volume in either litre or millilitre and use the same unit throughout the calculation to avoid errors. Carefully follow the steps to calculate the volume of the solution from the specific gravity given.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE