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Question: If x g of \(Na_2CO_3\) of 95% purity is required to neutralise 45.6 mL of 0.235 N acid, then the val...

If x g of Na2CO3Na_2CO_3 of 95% purity is required to neutralise 45.6 mL of 0.235 N acid, then the value of 10000x10000x is:
A.5978
B.6000
C.5500
D.none of the above

Explanation

Solution

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 Na2CO3\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 = milli equivalents of acid
Equivalents = NVEquivalents{\text{ }} = {\text{ }}NV (N = normality and V = volume)
It will be 0.235 × 45.6  0.235{\text{ }} \times {\text{ }}45.6\;
Now equivalents can also be written as WE\dfrac{W}{E}( where w is weight of substance Na2CO3\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 Na2CO3\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 Na have +2 charge and CO3\mathop {CO}\nolimits_3 has -2 charge thus n factor is 2 and its molar mass it 106) Now
E =1062E{\text{ }} = \dfrac{{106}}{2}
E = 53W53 × 1000 = 0.235 × 45.6E{\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= 0.5679 gram  W = {\text{ }}0.5679{\text{ }}gram\;
As 95 gram of Na2CO3\mathop {Na}\nolimits_2 \mathop {CO}\nolimits_3 is present in 100 gram of sample
So 1 gram will have =.10095\dfrac{{100}}{{95}}
Now 0.5679 will have = 10095× 0.5679 = 0.5978\dfrac{{100}}{{95}} \times {\text{ }}0.5679{\text{ }} = {\text{ }}0.5978
N now 10000 x will have = 0.5978 × 10000 = 5978 grams.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.