If Λc of NH4OH is 11.5Ω−1cm2mol−1, its degree of dissociation would be(Given λNH4+∞=73.4Ω−1cm2mol−1 and λOH∞=197.6Ω−1cm2mol−1.)

# If of ${\mathrm{NH}}_{4}\mathrm{OH}$ is $11.5{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{mol}}^{-1}$, its degree of dissociation would be(Given )

1. A

0.157

2. B

0.058

3. C

0.0424

4. D

0.0848

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### Solution:

degree of dissociation solution is  0.0424

$\alpha =\frac{\mathrm{\Lambda }}{{\mathrm{\Lambda }}^{\mathrm{\infty }}}=\frac{11.5}{\left(73.4+197.6\right)}=\frac{11.5}{271.0}=0.0424$.

When free ions carrying current are produced at a specific concentration and are dissociated from the solute fraction, this phenomenon is known as the degree of dissociation.

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