We are given two solutions: 750 $cm^3$ of 0.1 $mol \cdot dm^{-3}$ $NaCl$ and 250 $cm^3$ of 0.15 $mol \cdot dm^{-3}$ $Na_2SO_4$. These two solutions are mixed. We need to find the concentration of $Na^+$ in the final solution in $mol \cdot dm^{-3}$ and in $ppm$. We are given the atomic masses: $Na=23, Cl=35.5, O=16, S=32$.

1. Problem Description

We are given two solutions: 750 $cm^3$ of 0.1 $mol \cdot dm^{-3}$ $NaCl$ and 250 $cm^3$ of 0.15 $mol \cdot dm^{-3}$ $Na_2SO_4$. These two solutions are mixed. We need to find the concentration of $Na^+$ in the final solution in $mol \cdot dm^{-3}$ and in $ppm$. We are given the atomic masses: $Na=23, Cl=35.5, O=16, S=32$.

2. Solution Steps

First, we calculate the number of moles of $Na^+$ in each solution.

In $NaCl$ solution:

Volume = 750 $cm^3$ = 0.75 $dm^3$

Concentration of $NaCl$ = 0.1 $mol \cdot dm^{-3}$

Moles of $NaCl$ = Concentration $\times$ Volume = 0.1 $\times$ 0.75 = 0.075 moles

Since each mole of $NaCl$ gives one mole of $Na^+$, moles of $Na^+$ from $NaCl$ = 0.075 moles

In $Na_2SO_4$ solution:

Volume = 250 $cm^3$ = 0.25 $dm^3$

Concentration of $Na_2SO_4$ = 0.15 $mol \cdot dm^{-3}$

Moles of $Na_2SO_4$ = Concentration $\times$ Volume = 0.15 $\times$ 0.25 = 0.0375 moles

Since each mole of $Na_2SO_4$ gives two moles of $Na^+$, moles of $Na^+$ from $Na_2SO_4$ = 2 $\times$ 0.0375 = 0.075 moles

Total moles of $Na^+$ = 0.075 + 0.075 = 0.15 moles

Total volume of solution = 750 $cm^3$ + 250 $cm^3$ = 1000 $cm^3$ = 1 $dm^3$

Concentration of $Na^+$ = Total moles / Total volume = 0.15 moles / 1 $dm^3$ = 0.15 $mol \cdot dm^{-3}$

Now we need to calculate the concentration in $ppm$.

$ppm = \frac{\text{mass of solute}}{\text{mass of solution}} \times 10^6$

We can approximate the density of the solution as 1 g/mL (or 1 kg/L), so 1 L of solution has a mass of 1 kg = $10^6$ mg.

Moles of $Na^+$ = 0.15 moles

Molar mass of $Na$ = 23 g/mol

Mass of $Na^+$ = 0.15 moles $\times$ 23 g/mol = 3.45 g

Since the total volume is 1 $dm^3$ = 1 L, and we approximate the density of the solution as 1 kg/L, the mass of the solution is 1 kg = $10^6$ g.

$ppm = \frac{3.45 g}{1000 g} \times 10^6 = 3.45 \times 10^3 = 3450$

So the concentration of $Na^+$ is 0.15 $mol \cdot dm^{-3}$ and 3450 $ppm$.

3. Final Answer

0. 15, 3450