Question

500g of nacl was dissolved in a pool...

Answer 1

this will help u to solve the question

Answer 2

Answer: The correct answer is Option 1.

Explanation:

• To calculate the conductivity, we use the equation:

$\kappa=\frac{1}{R}\times G*$

where,

$\kappa$ = conductivity

R = resistance = $7600\Omega$

G* = Cell constant = $0.24cm^{-1}$

Putting values in above equation, we get:

$\kappa=\frac{0.24cm^{-1}}{7600\Omega}=3.16\times 10^{-5}\Omega ^{-1}cm^{-1}$

• To calculate the concentration from molar conductance, we use the equation:

$\Lambda_m=\frac{\kappa\times 1000}{M}$

where,

$\Lambda_m$ = molar conductance = $126.5\Omega cm^2mol^{-1}$

M = molarity of the solution = ?

Putting values in above equation, we get:

$126.5\Omega cm^2mol^{-1}=\frac{3.16\times 10^{-5}\Omega^{-1}cm^{-1}\times 1000(cm^3L^{-1}}{M}\\\\M=\frac{3.16\times 10^{-5}\times 1000}{126.5}=2.5\times 10^{-4}mol/L$

• To calculate the volume of solution, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of NaCl = 500 g

Molar mass of NaCl = 58.5 g/mol

Molarity of solution = $2.5\times 10^{-4}mol/L$

Putting values in above equation, we get:

$2.5\times 10^{-4}mol/L=\frac{500g}{58.5g/mol\times V}\\\\V=\frac{500}{58.5\times 2.5\times 10^{-4}}=0.034\times 10^6L$

Converting this into cubic decimeters, we use the conversion factor:

$1L=1dm^3$

So,

$0.034\times 10^4L=3.4\times 10^4dm^3$

Hence, the correct answer is Option 1.