Question

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1) Find the amount of 98% pure NA2CO3 required to prepare 5 litre of 1 Normality solution.


CLASS 9


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$<marquee > prabhudutt$

Answer 1

hey mate ,

       here is your answer :)

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Gram equivalent = Normality x Volume (in litre)

So gm equivalent required= 1 x 5 = 5

Equivalent mass = molar mass/ n- factor = 106/2= 53

Gm eq= mass of compound/ equivalent mass

So mass of compound = 5x 53= 265 g

Now 98g of Na2CO3 is present in 100g of sample

Therefore, amount of sample reqd. for 265g Na2CO3 = 100 x 265/98 = 270.408g