Solveeit Logo
Login

Question

Question: A big irregular shaped vessel contained water, conductivity of which was 2.56 × 10⁻³ S⁻¹m⁻¹. 585 g o...

A big irregular shaped vessel contained water, conductivity of which was 2.56 × 10⁻³ S⁻¹m⁻¹. 585 g of NaCl was then added to the water and conductivity after the addition of NaCl, was found to be 3.06 × 10⁻³ S⁻¹m⁻¹. The molar conductivity of NaCl at this concentration is 1.5 × 10⁻² S⁻¹m²mol⁻¹. The capacity of vessel if it is fulfilled with water, is

A

3 × 10⁴ L

B

3 × 10⁸ L

C

30 L

D

3 × 10⁵ L

Answer

3 × 10⁵ L

Explanation

Solution

  1. Calculate the conductivity due to NaCl: The conductivity of the solution is the sum of the conductivity of water and the conductivity contributed by NaCl. κNaCl=κsolution−κwater\kappa_{NaCl} = \kappa_{solution} - \kappa_{water} κNaCl=(3.06×10−3−2.56×10−3) S m−1=0.50×10−3 S m−1\kappa_{NaCl} = (3.06 \times 10^{-3} - 2.56 \times 10^{-3}) \text{ S m}^{-1} = 0.50 \times 10^{-3} \text{ S m}^{-1}

  2. Calculate the molar concentration of NaCl: The relationship between specific conductivity (κ\kappa), molar conductivity (Λm\Lambda_m), and molar concentration (cc) is κ=c×Λm\kappa = c \times \Lambda_m. c=κNaClΛmc = \frac{\kappa_{NaCl}}{\Lambda_m} c=0.50×10−3 S m−11.5×10−2 S m2 mol−1=0.501.5×10−1 mol m−3=13×10−1 mol m−3=130 mol m−3c = \frac{0.50 \times 10^{-3} \text{ S m}^{-1}}{1.5 \times 10^{-2} \text{ S m}^2 \text{ mol}^{-1}} = \frac{0.50}{1.5} \times 10^{-1} \text{ mol m}^{-3} = \frac{1}{3} \times 10^{-1} \text{ mol m}^{-3} = \frac{1}{30} \text{ mol m}^{-3}

  3. Calculate the number of moles of NaCl: The molar mass of NaCl is 23.0 g/mol (Na)+35.5 g/mol (Cl)=58.5 g/mol23.0 \text{ g/mol (Na)} + 35.5 \text{ g/mol (Cl)} = 58.5 \text{ g/mol}. nNaCl=Mass of NaClMolar mass of NaCl=585 g58.5 g/mol=10 moln_{NaCl} = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{585 \text{ g}}{58.5 \text{ g/mol}} = 10 \text{ mol}

  4. Calculate the volume of the solution: The molar concentration is moles of solute per unit volume of solution (c=n/Vc = n/V). Vsolution=nNaClcV_{solution} = \frac{n_{NaCl}}{c} Vsolution=10 mol130 mol m−3=10×30 m3=300 m3V_{solution} = \frac{10 \text{ mol}}{\frac{1}{30} \text{ mol m}^{-3}} = 10 \times 30 \text{ m}^3 = 300 \text{ m}^3

  5. Convert the volume to Liters: 1 m³ = 1000 L. Vsolution=300 m3×1000 L/m3=300,000 L=3×105 LV_{solution} = 300 \text{ m}^3 \times 1000 \text{ L/m}^3 = 300,000 \text{ L} = 3 \times 10^5 \text{ L} The capacity of the vessel is assumed to be equal to the volume of the solution formed.