If x g of $Na_2CO_3$ of 95% purity is required to neutralise 45.6 mL of 0.235 N acid, then the value of \[10000x\] is:
A.5978
B.6000
C.5500
D.none of the above
Answer
644.4k+ views
Hint: Neutralisation is defined as the reaction in which acid and base reacts to give salt and water. Acid neutralises the effect of base and base in turn neutralises the effect of acid. When neutralisation reaction occurs then the milli equivalents of acids and base become equal.
Complete step by step answer:
When neutralisation occurs:-
Milli equivalents of \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \]= milli equivalents of acid
\[Equivalents{\text{ }} = {\text{ }}NV\] (N = normality and V = volume)
It will be \[0.235{\text{ }} \times {\text{ }}45.6\;\]
Now equivalents can also be written as $\dfrac{W}{E}$( where w is weight of substance \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] and E is equivalent weight)
Equivalent weight will be M/n ( where M is molar mass and n is basicity or acidity or the valency as in in \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] Na have +2 charge and \[\mathop {CO}\nolimits_3 \]has -2 charge thus n factor is 2 and its molar mass it 106) Now
\[E{\text{ }} = \dfrac{{106}}{2}\]
\[E{\text{ }} = {\text{ }}\dfrac{{53W}}{{53}}{\text{ }} \times {\text{ }}1000{\text{ }} = {\text{ }}0.235{\text{ }} \times {\text{ }}45.6\] ( we multiply by thousands to convert milli to gram equivalents)
Now \[W = {\text{ }}0.5679{\text{ }}gram\;\]
As 95 gram of \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] is present in 100 gram of sample
So 1 gram will have =.$\dfrac{{100}}{{95}}$
Now 0.5679 will have = \[\dfrac{{100}}{{95}} \times {\text{ }}0.5679{\text{ }} = {\text{ }}0.5978\]
N now 10000 x will have = \[0.5978{\text{ }} \times {\text{ }}10000{\text{ }} = {\text{ }}5978{\text{ }}grams.\]
So ,our required answer is A i.e 5978.
Note:
Equivalent weight is usually defined as the molar mass divided by n factor. Neutralisation reaction is a type of double displacement reaction. In this reaction one product is always water. They are reversible reactions.
Complete step by step answer:
When neutralisation occurs:-
Milli equivalents of \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \]= milli equivalents of acid
\[Equivalents{\text{ }} = {\text{ }}NV\] (N = normality and V = volume)
It will be \[0.235{\text{ }} \times {\text{ }}45.6\;\]
Now equivalents can also be written as $\dfrac{W}{E}$( where w is weight of substance \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] and E is equivalent weight)
Equivalent weight will be M/n ( where M is molar mass and n is basicity or acidity or the valency as in in \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] Na have +2 charge and \[\mathop {CO}\nolimits_3 \]has -2 charge thus n factor is 2 and its molar mass it 106) Now
\[E{\text{ }} = \dfrac{{106}}{2}\]
\[E{\text{ }} = {\text{ }}\dfrac{{53W}}{{53}}{\text{ }} \times {\text{ }}1000{\text{ }} = {\text{ }}0.235{\text{ }} \times {\text{ }}45.6\] ( we multiply by thousands to convert milli to gram equivalents)
Now \[W = {\text{ }}0.5679{\text{ }}gram\;\]
As 95 gram of \[\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 \] is present in 100 gram of sample
So 1 gram will have =.$\dfrac{{100}}{{95}}$
Now 0.5679 will have = \[\dfrac{{100}}{{95}} \times {\text{ }}0.5679{\text{ }} = {\text{ }}0.5978\]
N now 10000 x will have = \[0.5978{\text{ }} \times {\text{ }}10000{\text{ }} = {\text{ }}5978{\text{ }}grams.\]
So ,our required answer is A i.e 5978.
Note:
Equivalent weight is usually defined as the molar mass divided by n factor. Neutralisation reaction is a type of double displacement reaction. In this reaction one product is always water. They are reversible reactions.
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