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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

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


 

Grade:11

1 Answers

Aman Bansal
592 Points
12 years ago

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

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