
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
511.5k+ 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.
Recently Updated Pages
Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain why it is said like that Mock drill is use class 11 social science CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Which one is a true fish A Jellyfish B Starfish C Dogfish class 11 biology CBSE
