How many ' Na' is required to prepare 10.6 grams of Na2Co3.
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Answer:
I hope it will be helpful to you.
Explanation:
You should add 5.299 grams of Sodium Bicarbonate in the 100 ml of solvent
you can calculate it using molarity relation
molarity = No. of moles / Liter of solution
Here, you have concentration , volume and molecular mass of Sodium bicarbonate
M = 0.1 , Vol= 0.100 Ltrs , lets assume number of moles = x (remember the units should be in same unit system)
substituting in above formula stated
we got x = 0.05 , now we need 0.05 molecules of sodium bicarbonate to be added in 100 ml so
now change the number of moles to grams
as we know 1 mole of Na2CO3 have molecular mass = 105.98
hence 0.05 moles will be = 0.05 x 105.98 = 5.299 grams.
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