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Equivalent, Molar Conductivity and Cell Constant

The molar con...

Question

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductiviy of $5.76 \times 10^{- 3} S c m^{- 1}$ at 298 K is

$2.88 c m^{2} / m o l$

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$11.52 S c m^{2} / m o l$

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$0.086 S c m^{2} / m o l$

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$28.8 c m^{2} / m o l$

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Solution

The correct option is B
$11.52 S c m^{2} / m o l$

$A_{m} = \frac{K \times 1000}{M o l a r i t y (M)}$
$= \frac{5.76 \times 10^{- 3} S c m^{- 1} \times 1000}{0.5 m o l c m^{- 3}} = 11.52 S c m^{2} m o l^{- 1}$

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Similar questions

Q. The molar conductivity of a

0.5 m o l / d m^{3}

solution of

A g N O_{3}

with electrolytic conductivity of

5.76 \times 10^{- 3} S c m^{- 1}

at 298 K is

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductiviy of $5.76 \times 10^{- 3} S c m^{- 1}$ at 298 K is

Q. The molar conductivity of a

0.5 m o l / d m^{3}

solution of

A g N O_{3}

with electrolytic conductivity of

5.76 \times 10^{- 3} S c m^{- 1}

298 K

is:

Q. The molar conductivity of a

0.5 m o l / d m^{3}

solution of

A g N O_{3}

with electrolytic conductivity of

5.76 \times 10^{- 3} S c m^{- 1}

at 298 K is

Q. Electrolytic conductivity of

0.3 M

solution of

K C I

298 K

3.72

10^{- 2} S c m^{- 1}

.Calculate its molar conductivity (

S c m^{2} m o l^{- 1}

)

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The molar conductivity of a 0.5 mol/dm3 solution of AgNO3 with electrolytic conductiviy of 5.76×10−3 S cm−1 at 298 K is

The correct option is B 11.52 S cm2/mol Am=K×1000Molarity (M) =5.76×10−3 S cm−1×10000.5 mol cm−3=11.52 S cm2 mol−1

The molar conductivity of a $0.5 m o l / d m^{3}$ solution of $A g N O_{3}$ with electrolytic conductiviy of $5.76 \times 10^{- 3} S c m^{- 1}$ at 298 K is

The correct option is B
$11.52 S c m^{2} / m o l$

$A_{m} = \frac{K \times 1000}{M o l a r i t y (M)}$
$= \frac{5.76 \times 10^{- 3} S c m^{- 1} \times 1000}{0.5 m o l c m^{- 3}} = 11.52 S c m^{2} m o l^{- 1}$