Question

15. 500 g of NaCl was dissolved in a pool which was
thoroughly stirred and the sample of this solution
gave resistance of 7600 Q when placed in a
conductivity cell of cell constant 0.24 cm-. Then,
volume of water in the pool will be (Neglect
conductance of water)
Given: Molar conductance of NaCl (at that
concentration) is 126.5 2-1 cm mol .
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Answer 1

Answer:

Conductance of solution due to NaCl , G = 1Resistance = 17600 Omega = 1.31 ×10−4 Omega−1

Conductivity of NaCl = G K

= (1.31 ×10−3×0.24 cm−1)

= 3.15 ×10−5 Omega−1 cm−1

Concentartion of NaCl in the swimming pool:

c = κ∧m

= 3.15 ×10−5126.5 = 2.49 ×10−7 mol cm−3

= 2.49 ×10−4 mol dm−3

Mass of NaCl in 1 dm3 solution = (2.49 ×10−4 mol dm−3 ×58.5 g mol−1 ) = 0.0145 g ( Molar mass of NaCl = 58.5 g mol-1)

Total volume of water in the pool containing 500 g NaCl = (1 dm30.0145 g ) × 500 g = 3.448 ×103 dm3

Since 1 dm3 = 1 L

So, volume in L = 3.448 x 103 L = 3448 L