1. 5 gm of K2SO4 were dissolved in 250 ml of solution . the volume of this solution that should be used so thst 1.2 gm of baso4 may be precipitated from bacl3 is......... ( mol. mass of K2SO4 = 174 and baso4 = 233)2. the equi.weight of NAHCO3 is..... and of so2 is........3. a sample of clay was partially dried and then found to contain 50% silica and 7% water . the originsl clay contained 12% water.what is the percentage of silica in in original solution?4.a certain compound has a molecular formula X4O6 .if 10 gm of this compound has 5.72 gms of X ,,then atomic mass of X is.....Thank you

Dear Siddhant,

For Na2CO3 there are two Na+ atoms to react with acids. Therefore in case of Na2CO3 equivalent weight is Molecular weight / 2.
In the case of NaHCO3 there is only one Na+ to react with acids. (You can imagine that one Na of Na2CO3 has been used by an acid) Therefore equivalent weight is Molecular weight / 1, or molecular weight itself. In case of these compounds it is Na to decide its acidity, not the Proton

Best Of luck

Plz Approve the answer...!!!!

Cracking IIT just got more exciting,It’s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and winexciting gifts by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple click here to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal

Askiitian Expert