The specific conductance of a 0.002 N solution of Na2SO4 is 2.6 × 10-2 Sm-1 at 25°C and it rises to 7 ×10-2 Sm-1 on saturating it with CaSO4. The molar conductance at infinite dilution of Na+ and Ca2+ ions are 50 × 10-4 and 12 × 10-4 S m2 mol-1 respectively. Correct statements are

(1) Specific conductance due to CaSO4=7×10−2−2.6×10−2=4.4×10−2Sm−1 Assuming Na2SO4 to be very dilute, μ0Na2SO4=kc=2.6×10−2Sm−10.001mollitre−1×103lm−3⏟ =2.6×10−2Sm2mol-1(2) μ0SO42−=0.026−2×5×10−3=1.6×10−2Sm2mol−1(3) μ0CaSO4=1.2×10−2+1.6×10−2=2.8×10−2Sm2mol−1 Solubility of CaSO4=4.4×10−22.8×10−2Sm−1Sm2mol−1=1.57molm−3=1.57×10−3mollitre−1(4) KSPCaSO4=1.57×10−32=2.465×10−6

The specific conductance of a 0.002 N solution of Na2SO4 is 2.6 × 10-2 Sm-1 at 25°C and it rises to 7 ×10-2 Sm-1 on saturating it with CaSO4. The molar conductance at infinite dilution of Na+ and Ca2+ ions are 50 × 10-4 and 12 × 10-4 S m2 mol-1 respectively. Correct statements are

(1) Specific conductance due to CaSO4=7×10−2−2.6×10−2=4.4×10−2Sm−1 Assuming Na2SO4 to be very dilute, μ0Na2SO4=kc=2.6×10−2Sm−10.001mollitre−1×103lm−3⏟ =2.6×10−2Sm2mol-1(2) μ0SO42−=0.026−2×5×10−3=1.6×10−2Sm2mol−1(3) μ0CaSO4=1.2×10−2+1.6×10−2=2.8×10−2Sm2mol−1 Solubility of CaSO4=4.4×10−22.8×10−2Sm−1Sm2mol−1=1.57molm−3=1.57×10−3mollitre−1(4) KSPCaSO4=1.57×10−32=2.465×10−6

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The specific conductance of a 0.002 N solution of Na₂SO₄ is 2.6 × 10^-2 Sm^-1 at 25°C and it rises to 7 ×10^-2 Sm^-1on saturating it with CaSO₄. The molar conductance at infinite dilution of Na⁺ and Ca²⁺ ions are 50 × 10^-4 and 12 × 10^-4 S m² mol^-1 respectively. Correct statements are

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$The specific conductance (in {Sm}^{- 1}) of {CaSO}_{4} solution is 1.8 \times 10^{- 2}$

$The molar ionic conductance of {SO}_{4}^{2 -} (in {Sm}^{2} {mol}^{- 1}) is 0.016$

$The solubility of {CaSO}_{4} (in {moldm}^{- 3}) is 1.57 \times 10^{- 3}$

$The K_{sp} of {CaSO}_{4} is ({mol}^{2} {dm}^{- 6}) is 2.465 \times 10^{- 3} .$

answer is B, C.

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Detailed Solution

(1) $Specific conductance due to {CaSO}_{4} = 7 \times 10^{- 2} - 2.6 \times 10^{- 2} = 4.4 \times 10^{- 2} {Sm}^{- 1}$

$Assuming {Na}_{2} {SO}_{4} to be very dilute,$

$μ_{0} ({Na}_{2} {SO}_{4}) = \underset{}{\underset{⏟}{(\frac{k}{c} = \frac{2.6 \times 10^{- 2} ({Sm}^{- 1})}{0.001 ({mollitre}^{- 1} \times 10^{3} {lm}^{- 3})})}}$

$= 2.6 \times 10^{- 2} {Sm}^{2} mol$ ^-1

(2) $μ_{0} ({SO}_{4}^{2 -}) = 0.026 - (2 \times 5 \times 10^{- 3}) = 1.6 \times 10^{- 2} {Sm}^{2} {mol}^{- 1}$

(3) $μ_{0} ({CaSO}_{4}) = 1.2 \times 10^{- 2} + 1.6 \times 10^{- 2} = 2.8 \times 10^{- 2} {Sm}^{2} {mol}^{- 1}$

$\begin{aligned} Solubility of {CaSO}_{4} = \frac{4.4 \times 10^{- 2}}{2.8 \times 10^{- 2}} (\frac{{Sm}^{- 1}}{{Sm}^{2} {mol}^{- 1}}) = 1.57 {molm}^{- 3} \\ = 1.57 \times 10^{- 3} mol {litre}^{- 1} \end{aligned}$