Equivalent conductance of saturated solution of BaSO 4 is 200 ohm -1 cm 2 eqv -1 and specific conductance is 4 - 10-5 ohm -1 cm -1 - K sp of BaSO 4 x - 10- y M 2 in scientific notation- The value of x - y will be-

Given : Equivalent conductance nbsp; = 200 ohm^ 1 cm^2 eqv^ 1 Specific conductance (κ nbsp;) = 4 × 10^ 5 ohm^ 1 cm^ 1 we know, nbsp; equivalent conductan ...

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Standard XII

Chemistry

Conductance and Conductivity

Equivalent co...

Question

Equivalent conductance of saturated solution of $B a S O_{4}$ is 200 $o h m^{- 1} c m^{2} e q v^{- 1}$ and specific conductance is $4 \times 10^{- 5} o h m^{- 1} c m^{- 1} . K_{s p}$ of $B a S O_{4} x \times 10^{- y} M^{2}$ in scientific notation. The value of (x + y) will be:

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Solution

Given : Equivalent conductance = 200 $o h m^{- 1} c m^{2} e q v^{- 1}$
Specific conductance $(κ) = 4 \times 10^{- 5} o h m^{- 1} c m^{- 1}$
we know,
$e q u i v a l e n t c o n d u c t a n c e = \frac{κ \times 1000}{N o r m a l i t y}$

$N o r m a l i t y = \frac{4 \times 10^{- 5} \times 1000}{200} = 2 \times 10^{- 4}$
(n-factor of $B a S O_{4} = 2$ )
Molarity $= 10^{- 4}$
Now, $B a S O_{4} \to B a^{2 +} + S O_{4}^{2 -}$
$K_{s p} = [B a^{2 +}] [S O_{4}^{2 -}] = [C] . [C]$
$= 10^{- 4} \times 10^{- 4} = 1 \times 10^{- 8} M^{2}$
$x = 1, y = 8, S o x + y = 9$

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Equivalent conductance of saturated solution of BaSO4 is 200 ohm−1 cm2 eqv−1 and specific conductance is 4×10−5 ohm−1 cm−1. Ksp of BaSO4 x ×10−y M2 in scientific notation. The value of (x + y) will be:

Equivalent conductance of saturated solution of $B a S O_{4}$ is 200 $o h m^{- 1} c m^{2} e q v^{- 1}$ and specific conductance is $4 \times 10^{- 5} o h m^{- 1} c m^{- 1} . K_{s p}$ of $B a S O_{4} x \times 10^{- y} M^{2}$ in scientific notation. The value of (x + y) will be: