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Grade: 12
        please please answer this question please please sir
how many ml of 0.150 M Na2CrO will be required to oxidize 40 ml of 0.5 M  Na2S2O3
Cr2O42-  +  S2O32-  ------> Cr(OH)4+ So42-
a. 225 ml
b. 335 ml
c. 455 ml
d.555 ml  
one year ago

Answers : (1)

Sahani Kumar
99 Points
							
 
 
 
Cr2O42-  +  S2O32-  ------> Cr(OH)4+ So42-
in the above equation sulphur (S) is being oxidised from S2+  to S6+ as follows
S2+ -------------->S6+ + 4e-         So n1=4
in the same way Cr is being reduced from Cr3+ to Cr2+ as follows  
Cr3+ + 1e- --------------->Cr2+            So n2= 1
Now we know that 
n1 * M1 * V1 = n2 * M2 * V2
4 * 0.5 * 40 = 1 * 0.150 * V2
V2 = (4 * 0.5 * 40)/(1 * 0.150) = 533 ml ~ 555 ml(which is almost nearer to the value 555 ml. )
Hence according to me the correct answer is 555 ml please let me know if this answer is correct. 
Regards 
Askiitian member
 
 
 
 
one year ago
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