At 291 K, saturated solution of BaSO4 was found to have a specific conductivity of 3.648×10-6ohm-1 cm-1 and that of water being used is 1.25×10-6ohm-1 cm-1. If the ionic conductance's of Ba2+ and SO42- are 110 and 136.6ohm-1 cm2 mol-1 respectively, the solubility of BaSO4 at 291 K will be [Atomic masses ofBa-137,S-32,0-16

AP EAMCET · Chemistry · Electrochemistry

At $291 K$ , saturated solution of ${BaSO}_{4}$ was found to have a specific conductivity of $3.648 \times 10^{- 6} {ohm}^{- 1} {cm}^{- 1}$ and that of water being used is $1.25 \times 10^{- 6} {ohm}^{- 1} {cm}^{- 1}$ . If the ionic conductance's of ${Ba}^{2 +}$ and ${SO}_{4}^{2 -}$ are $110$ and $136.6 {ohm}^{- 1} {cm}^{2} {mol}^{- 1}$ respectively, the solubility of ${BaSO}_{4}$ at $291 K$ will be [Atomic masses of $Ba - 137, S - 32, 0 - 16]$

A $1 .435 \times 10^{- 3} {gL}^{- 1}$
B $2 .266 \times 10^{- 3} {gL}^{- 1}$
C $2 .843 \times 10^{- 3} {gL}^{- 1}$
D $1 .768 \times 10^{- 3} {gL}^{- 1}$

Verified Solution

Answer & Solution

Correct Answer

(B) $2 .266 \times 10^{- 3} {gL}^{- 1}$

Step-by-step Solution

Detailed explanation

According to Kohlrausch's law λ°mBaSO4 = λ°mBa2++λ°mSO42-λ°mBaSO4 = 110+136.6λ°mBaSO4 = 246.6 ohm-1cm-1 Now, the solubility product of BaSO4 can be determined by:…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match the following elements, given as hints to their first, second ionization enthalpies and
electron gain enthalpies.

List- $I$	List- $II$
$i)$ Most reactive non-metal	$A)$ $∆ H_{1} = 419, ∆ H_{2} = 3051, ∆_{Eg} H = - 48$
$ii)$ Most reactive metal	$B)$ $∆ H_{1} = 1681, ∆ H_{2} = 3374, ∆_{Eg} H = - 328$
$iii)$ Least reactive element	$C)$ $∆ H_{1} = 738, ∆ H_{2} = 1451, ∆_{Eg} H = - 40$
$iv)$ Metal forming binary halide	$D)$ $∆ H_{1} = 3272, ∆ H_{2} = 5251, ∆_{Eg} H = + 48$

AP EAMCET 2020 Easy

Match the complexes in list-I with their hybridization in list-II.
$\begin{array}{|l|l|l|l|}\hline & \text{List-I (Complex)} & & \text{List-II (Hybridization)} \\ \hline \text{A.} & \mathrm{Ni}(\mathrm{CO})_4 & I & s p^3 d^2 \\ \hline \text{B.} & {\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}} & II & d^2 s p^3 \\ \hline \text{C.} & {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right]^{3+}} & III & d s p^2 \\ \hline \text{D.} & {\left[\mathrm{CoF}_6\right]^{3-}} & IV & s p^3 \\ \hline\end{array}$

AP EAMCET 2024 Easy

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