If Λeq∞(NaCl),Λeq∞(KCl) and Λeq∞K2SO4 are 123.7, 147.0 and 152.1Ω−1cm2eq−1, then the Λeq∞Na2SO4 would be

# If , then the ${\mathrm{\Lambda }}_{\mathrm{eq}}^{\mathrm{\infty }}\left({\mathrm{Na}}_{2}{\mathrm{SO}}_{4}\right)$ would be

1. A

$128.8{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}$

2. B

$257.6{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}$

3. C

$105.5{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}$

4. D

$118.6{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}$

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

${\mathrm{\Lambda }}_{\mathrm{eq}}^{\mathrm{\infty }}\left({\mathrm{Na}}_{2}{\mathrm{SO}}_{4}\right)={\mathrm{\Lambda }}_{\mathrm{eq}}^{\mathrm{\infty }}\left(\mathrm{NaCl}\right)+{\mathrm{\Lambda }}_{\mathrm{eq}}^{\mathrm{\infty }}\left({\mathrm{K}}_{2}{\mathrm{SO}}_{4}\right)-{\mathrm{\Lambda }}_{\mathrm{eq}}^{\mathrm{\infty }}\left(\mathrm{KCl}\right)=\left(123.7+152.1-147.0\right){\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}=128.8{\mathrm{\Omega }}^{-1}{\mathrm{cm}}^{2}{\mathrm{eq}}^{-1}$

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)