Question

Find the amount of 98% pure Na2Co3 required to prepare 5 liters of 1 N solution plz answer this

Answer 1

Answer:

270.408g

Explanation:

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

Answer 2

Answer:

270.408g of sodium carbonate