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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Chemistry
Chapter
Electrochemistry

BITSAT 2009 · Chemistry · Electrochemistry

The conductivity of a saturated solution of BaSO_ 4 is 3.06 × 10^...

The conductivity of a saturated solution of BaSO4\mathrm{BaSO}_{4} is 3.06×10−6ohm−1cm−13.06 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{ cm}^{-1} and its equivalent conductance is 1.53ohm−1cm21.53 \mathrm{ohm}^{-1} \mathrm{ cm}^{2} equiv ^{-1}.TheKsp. The \mathrm{K}_{\mathrm{sp}} for BaSO4\mathrm{BaSO}_{4} will be

Options

  1. A4×10−124 \times 10^{-12}
  2. B2.5×10−92.5 \times 10^{-9}
  3. C2.5×10−132.5 \times 10^{-13}
  4. D4×10−64 \times 10^{-6}

Correct answer

D. 4×10−64 \times 10^{-6}

Step-by-step solution

Solublity = conductivity ×1000Λeq =\frac{\text{ conductivity } \times 1000}{\Lambda_{\text{eq }}}=3.06×10−6×10001.53=2×10−3Ksp =S2=4×10−6\begin{array}{l} =\frac{3.06 \times 10^{-6} \times 1000}{1.53}=2 \times10^{-3} \\ K_{\text{sp }}=\mathrm{S}^{2}=4 \times10^{-6} \end{array}
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