At $291 \ K$,a saturated solution of $BaSO_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and that of the water used is $1.25 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$. If the ionic conductances of $Ba^{2+}$ and $SO_4^{2-}$ are $110$ and $136.6 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ respectively,the solubility of $BaSO_4$ at $291 \ K$ will be (Atomic masses: $Ba=137, S=32, O=16$)

  • A
    $1.435 \times 10^{-3} \ g \ L^{-1}$
  • B
    $2.266 \times 10^{-3} \ g \ L^{-1}$
  • C
    $2.843 \times 10^{-3} \ g \ L^{-1}$
  • D
    $1.768 \times 10^{-3} \ g \ L^{-1}$

Explore More

Similar Questions

Write the formula to calculate conductivity by using a conductivity cell.

The resistance of $0.01 \ N$ $NaCl$ solution at $25 \ ^oC$ is $200 \ \Omega$. The cell constant of the conductivity cell is $1 \ cm^{-1}$. The equivalent conductance is:

The variation of molar conductivity with concentration of an electrolyte $X$ in aqueous solution is shown in the given figure.
The electrolyte $X$ is :

At $25^{\circ} C$,the molar conductances at infinite dilution for the strong electrolytes $NaOH$,$NaCl$ and $BaCl_2$ are $248 \times 10^{-4}$,$126 \times 10^{-4}$ and $280 \times 10^{-4} \ S \ m^2 \ mol^{-1}$ respectively. The value of $\lambda_m^{\circ} Ba(OH)_2$ in $S \ m^2 \ mol^{-1}$ is:

$A$ conductivity cell has two electrodes $18 \ mm$ apart and having cross-sectional area $2.0 \ cm^2$. What is the value of the cell constant (in $cm^{-1}$)?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo