The values of observed and calculated molecular weights of silver nitrate are `92.64` and `170` respectively.The degree of dissociation of silver nitrate is:
A. 0.6
B. 0.835
C. 0.467
D. 0.6023

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Correct Answer - B
`alpha = (i-1)/(n^(l) - 1)`
For `AgNO_(3), n^(l) = 1 :. alpha = (i-1)/(2-1) :. alpha = i-1``"But" i="normal molar mass."/"observed molar mass."`
`:. alpha = 170/92.64 - 1 = 0.835`
`:. % alpha = 0.835 xx 100 = 83.5 %`

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