The volume of 2M NaOH required to convert 12 grams of NaH2PO4 [M=120gm mol-1] into Na3PO4 is

(A) 20 ml

(B) 80 ml

(C) 100 ml

(D) 200 ml

Concept Videos :-

#2 | 15th Group: Properties: II
#3 | 15th Group: Ammonia

Concept Questions :-

group 15 ,preparation and properties

The reaction is

NaH2PO4 + 2NaOH  Na3PO4 +2H2O n120g              2×40=80g(1 mole)         (2 moles)120g NaH2PO4 requires 80g NaOH12g NaH2PO4requires 8 g NaOH80g NaOH is present in 1 litre i.e. 1000 ml of 2M NaOH8g NaOH is present in 100080×8=100ml of 2M NaOH

 

 

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