order solutions by conductivity

The conductivity of ionic solutions depends on the concentration and charge of the ions. Higher concentration and higher charges lead to greater conductivity. $\text{Na}_2\text{SO}_4$ dissociates into 3 ions ($2\text{Na}^+ + \text{SO}_4^{2-}$), $\text{KCl}$ into 2 ions ($1\text{K}^+ + 1\text{Cl}^-$), and $\text{LiCl}$ into 2 ions ($1\text{Li}^+ + 1\text{Cl}^-$). $\text{NaCl}$ also dissociates into 2 ions ($1\text{Na}^+ + 1\text{Cl}^-$). Comparing the effective ion concentrations: $1.0 M \text{Na}_2\text{SO}_4 \rightarrow 3.0 M$ ions, $1.2 M \text{KCl} \rightarrow 2.4 M$ ions, $0.75 M \text{LiCl} \rightarrow 1.5 M$ ions, and $1.0 M \text{NaCl} \rightarrow 2.0 M$ ions. Therefore, the order of increasing conductivity is $0.75 M \text{LiCl} < 1.0 M \text{NaCl} < 1.2 M \text{KCl} < 1.0 M \text{Na}_2\text{SO}_4$. The circles are labeled 1, 2, 3, and 4, and the solutions are listed above the circles and to the right of circle 3. Assuming the circles represent positions and the solutions are to be placed in order of increasing conductivity, circle 1 should contain $0.75 M \text{LiCl}$, circle 2 should contain $1.0 M \text{NaCl}$, circle 3 should contain $1.2 M \text{KCl}$, and the solution labeled 4 ($1.0 M \text{NaCl}$) is already placed. However, the image shows 3 circles and then a box labeled 4. The arrow indicates increasing conductivity. The question asks to order the given solutions. The solutions are $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$, and $1.0 M \text{NaCl}$. The circles are numbered 1, 2, 3, and there is a box labeled 4. The prompt asks to solve the problem and give only the answer. The problem is to order these solutions by conductivity. The order of increasing conductivity is $0.75 M \text{LiCl} < 1.0 M \text{NaCl} < 1.2 M \text{KCl} < 1.0 M \text{Na}_2\text{SO}_4$. If the circles and box represent slots to place these solutions in order of increasing conductivity, then circle 1 should be $0.75 M \text{LiCl}$, circle 2 should be $1.0 M \text{NaCl}$, circle 3 should be $1.2 M \text{KCl}$, and box 4 should be $1.0 M \text{Na}_2\text{SO}_4$. The image shows the solutions $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$ at the top, and $1.0 M \text{NaCl}$ at the right. The circles are labeled 1, 2, 3. The text below indicates "Least conductive" and "Increasing Conductivity". This implies that the solutions should be placed in the circles and box in order of increasing conductivity. Thus, circle 1 should be $0.75 M \text{LiCl}$, circle 2 should be $1.0 M \text{NaCl}$, circle 3 should be $1.2 M \text{KCl}$, and the solution $1.0 M \text{Na}_2\text{SO}_4$ should be placed in the position corresponding to the highest conductivity. The box labeled 4 is to the right of circle 3. Given the arrow pointing down and the labels "Increasing Conductivity" and "Least conductive", it is implied that the solutions should be ordered from left to right or top to bottom. The solutions at the top are $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$. The solution $1.0 M \text{NaCl}$ is to the right of circle 3. The circles are numbered 1, 2, 3. The question is to order the solutions. The order of increasing conductivity is $0.75 M \text{LiCl} < 1.0 M \text{NaCl} < 1.2 M \text{KCl} < 1.0 M \text{Na}_2\text{SO}_4$. If the circles and box represent the order, then 1 should be $0.75 M \text{LiCl}$, 2 should be $1.0 M \text{NaCl}$, 3 should be $1.2 M \text{KCl}$, and the remaining solution $1.0 M \text{Na}_2\text{SO}_4$ should be placed in the position of highest conductivity. The image shows the solutions $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$ at the top, and $1.0 M \text{NaCl}$ to the right of circle 3. The circles are labeled 1, 2, 3. The text indicates "Least conductive" and "Increasing Conductivity". This implies that the solutions should be placed in the numbered positions in increasing order of conductivity. Therefore, position 1 should be $0.75 M \text{LiCl}$, position 2 should be $1.0 M \text{NaCl}$, position 3 should be $1.2 M \text{KCl}$, and the remaining solution $1.0 M \text{Na}_2\text{SO}_4$ should be placed in the position of highest conductivity. The image shows the solutions $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$ at the top. The circles are numbered 1, 2, 3. The solution $1.0 M \text{NaCl}$ is to the right of circle 3. The arrow indicates increasing conductivity. The text "Least conductive" is below the circles. This means the order of conductivity is from left to right. So, circle 1 should be the least conductive, and the subsequent positions should be increasingly conductive. The solutions are $1.0 M \text{Na}_2\text{SO}_4$, $1.2 M \text{KCl}$, $0.75 M \text{LiCl}$, $1.0 M \text{NaCl}$. Their effective ion concentrations are approximately 3.0 M, 2.4 M, 1.5 M, and 2.0 M, respectively. Thus, the order of increasing conductivity is $0.75 M \text{LiCl} < 1.0 M \t...