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Que. 1 The milli moles of Borax (Na 2 B 4 O 7 .10H 2 O) are required to neutralize 25 mL each of 0.2 M H 2 SO 4 and H 3 PO 3 respectively are ?? que 2 An ore contains 2.5% of the mineral argentite, Ag 2 S, by mass. How many grams of this ore would have to be processed in order to obtain 0.675g of pure solid silver, Ag? Que 3 in a titration, 60 mL of Phosphoric acid (H 3 PO 4 ) completely neutralizes 750 mL of Sodium carbonate solution containing 3.18 gm of Na 2 CO 3 in one litre of solution. Thus the molarity of Phosphoric acid is??

 


Que. 1


The milli moles of Borax (Na2B4O7.10H2O) are required to neutralize 25 mL each of 0.2 M H2SO4 and H3PO3 respectively are ??


que 2



An ore contains 2.5% of the mineral argentite, Ag2S, by mass. How many grams of this ore would have to be processed in order to obtain 0.675g of pure solid silver, Ag?


 


Que 3



in a titration, 60 mL of Phosphoric acid (H3PO4) completely neutralizes 750 mL of Sodium carbonate solution containing 3.18 gm of Na2CO3


 


in one litre of solution. Thus the molarity of Phosphoric acid is??




 

Grade:11

1 Answers

Aman Bansal
592 Points
9 years ago

Dear Aman,

1)calculate milliequivalents of h3po2 by multiplying volume and normality
No of milliequivalents = 100 * 0.2 = 20 milllieqs
millieqs of mg(oh)2 is 80 * 0.2 = 16millieqs
so 4 millieqs are left for nuetralization
thus we require 4 millieqs of koh
so 4 = vol * normality 
so 4 = volume * 0.25 
volume = 16ml
2)in ques 2 calculate the milli moles of h2so4 by multiplying molarity and volume ( in ml)
similarly for h3po3
then apply stoichiometry 
to the eqaution Na2B4O7 + 2HCl + 5H2O --> 2NaCl + 4H3BO3
millimoles of hcl = 25*0.2 = 5 millimoles
acc to equation 1mole of borax react with 2 moles of hcl
so 2.5 moles of borax react with 5 moles of hcl
similarly for h3po3
3)let the mass of ore be x
then 2.5 * x /100 = molecular wt of ag2s = 248
so x = 9920
so 9920 gram of ore yields 216 grams of silver ( ag2)
so 31 grams of ore with 0.675 gram of ag

Best Of luck

Plz Approve the answer...!!!!

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Thanks

Aman Bansal

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