Answer
Verified
399.6k+ views
Hint: Normality is the number of equivalents of solute in one litre of solution. The strength of the solution is determined using normality. Equivalent weight can be calculated by dividing molecular weight by valency.
Given data:
Normality of oxalic acid, \[{\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}\]
Volume of oxalic acid, \[{\text{V}} = 250{\text{mL}}\]
Complete step by step solution:
Oxalic acid is having the formula\[{\text{COOH}} - {\text{COOH}}.2{{\text{H}}_2}{\text{O}}\]. It is a dibasic acid since it can donate two protons. Therefore its n-factor or valency is 2.
Normality is expressed as the number of gram equivalents/one litre of solution. Its unit is generally expressed as N. Number of gram equivalents indicates the number of grams of reactive species in a compound.
Given that normality \[{\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}\]
Volume of acid, \[{\text{V}} = 250{\text{mL}}\]
Molecular mass of oxalic acid, ${\text{M}} = 126$
Normality=Number of gram equivalent in volume (litre)
Equivalent weight of oxalic acid, \[{\text{E}} = \dfrac{{\text{M}}}{{\text{n}}}\], where \[{\text{M}}\]is the molecular weight of oxalic acid.
\[{\text{n}}\]is the n-factor or the valency which is equal to 2.
\[{\text{E}} = \dfrac{{126{\text{g}}}}{2} = 63{\text{g}}\]
Number of gram equivalence\[{\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{\text{E}}}\], \[{\text{w}}\]is the given weight of solute which we have to calculate here.
\[{\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{63{\text{g}}}}\]
Now normality can be calculated using these values.
\[{\text{N}} = \dfrac{{{\text{G}}.{\text{eq}}}}{{\text{V}}}\]
Substitute the value of number of gram equivalence in the above equation, it becomes
\[{\text{N}} = \dfrac{{\dfrac{{\text{w}}}{{63{\text{g}}}}}}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}\]
Given that \[{\text{N}} = 0.1{\text{N}}\]in one litre.
i.e., \[{\text{N}} = \dfrac{{0.1}}{{1000{\text{mL}}}}\]
Substituting the value of normality, we get
\[\dfrac{{0.1}}{{1000{\text{mL}}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}\]
\[{\text{w}} = \dfrac{{0.1 \times 63{\text{g}} \times 250{\text{mL}}}}{{1000{\text{mL}}}}\]
Now we get the amount of solute to be taken.
\[{\text{w}} = \dfrac{{1575{\text{gmL}}}}{{1000{\text{mL}}}} = 1.575{\text{g}}\]
Hence option A is correct.
\[1.575{\text{g}}\]of oxalic acid has to be taken to prepare \[250{\text{mL}}\]of \[\dfrac{{\text{N}}}{{10}}\]solution.
Additional information:
When oxalic acid is dissolved in water it dissociates into two protons. Therefore its n-factor tends to be two. Normality measures the concentration of solution. It is also known as the equivalent concentration of solution.
Note: Normality and molarity have an important relation. Normality is the product of molarity and acidity or basicity. Acidity is the number of hydroxyl ions the molecule can give. Basicity is the number of protons the molecule can give.
Given data:
Normality of oxalic acid, \[{\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}\]
Volume of oxalic acid, \[{\text{V}} = 250{\text{mL}}\]
Complete step by step solution:
Oxalic acid is having the formula\[{\text{COOH}} - {\text{COOH}}.2{{\text{H}}_2}{\text{O}}\]. It is a dibasic acid since it can donate two protons. Therefore its n-factor or valency is 2.
Normality is expressed as the number of gram equivalents/one litre of solution. Its unit is generally expressed as N. Number of gram equivalents indicates the number of grams of reactive species in a compound.
Given that normality \[{\text{N}} = \dfrac{{\text{N}}}{{10}} = \dfrac{1}{{10}}{\text{N}} = 0.1{\text{N}}\]
Volume of acid, \[{\text{V}} = 250{\text{mL}}\]
Molecular mass of oxalic acid, ${\text{M}} = 126$
Normality=Number of gram equivalent in volume (litre)
Equivalent weight of oxalic acid, \[{\text{E}} = \dfrac{{\text{M}}}{{\text{n}}}\], where \[{\text{M}}\]is the molecular weight of oxalic acid.
\[{\text{n}}\]is the n-factor or the valency which is equal to 2.
\[{\text{E}} = \dfrac{{126{\text{g}}}}{2} = 63{\text{g}}\]
Number of gram equivalence\[{\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{\text{E}}}\], \[{\text{w}}\]is the given weight of solute which we have to calculate here.
\[{\text{G}}.{\text{eq}} = \dfrac{{\text{w}}}{{63{\text{g}}}}\]
Now normality can be calculated using these values.
\[{\text{N}} = \dfrac{{{\text{G}}.{\text{eq}}}}{{\text{V}}}\]
Substitute the value of number of gram equivalence in the above equation, it becomes
\[{\text{N}} = \dfrac{{\dfrac{{\text{w}}}{{63{\text{g}}}}}}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{\text{V}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}\]
Given that \[{\text{N}} = 0.1{\text{N}}\]in one litre.
i.e., \[{\text{N}} = \dfrac{{0.1}}{{1000{\text{mL}}}}\]
Substituting the value of normality, we get
\[\dfrac{{0.1}}{{1000{\text{mL}}}} = \dfrac{{\text{w}}}{{63{\text{g}}}} \times \dfrac{1}{{250{\text{mL}}}}\]
\[{\text{w}} = \dfrac{{0.1 \times 63{\text{g}} \times 250{\text{mL}}}}{{1000{\text{mL}}}}\]
Now we get the amount of solute to be taken.
\[{\text{w}} = \dfrac{{1575{\text{gmL}}}}{{1000{\text{mL}}}} = 1.575{\text{g}}\]
Hence option A is correct.
\[1.575{\text{g}}\]of oxalic acid has to be taken to prepare \[250{\text{mL}}\]of \[\dfrac{{\text{N}}}{{10}}\]solution.
Additional information:
When oxalic acid is dissolved in water it dissociates into two protons. Therefore its n-factor tends to be two. Normality measures the concentration of solution. It is also known as the equivalent concentration of solution.
Note: Normality and molarity have an important relation. Normality is the product of molarity and acidity or basicity. Acidity is the number of hydroxyl ions the molecule can give. Basicity is the number of protons the molecule can give.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE