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Q.

what is the molality of dilute aqueous 0.02N H3PO4

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a


0.0050

b

0.0067

c

0.0200

d

0.00330

answer is D.

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Detailed Solution

Normality, N = 0.02
It is known that N=n1×M
N=n1×Number of moles/ Volume of solution(L)
For H3PO4, n=3 
Therefore, N=3×Number of moles /Volume of solution(L)
 Number of moles/ Volume of solution(kg)=molality of solution
, N3=0.02/3=0.0067 molal.

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