Calculate the number of moles of the indicated ion present in each of the following solutions. Na+ ion in 1.20 L of 0.183 M Na2SO4 solution = ?mol
mols Na2SO4 = M Na2SO4 x L Na2SO4.
Then Na^+ will be twice that since there are two moles Na^+2 for each mole of Na2SO4.
Well, let's do the math. Na2SO4 is a salt that dissociates into three ions when it dissolves in water - 2 Na+ ions and 1 SO4^2- ion. Since the concentration of Na2SO4 is 0.183 M, this means that there are 0.183 moles of Na2SO4 per liter.
Since Na2SO4 contains 2 Na+ ions, the number of moles of Na+ ions in the solution can be calculated by multiplying the concentration of Na2SO4 by the number of Na+ ions per formula unit.
0.183 M Na2SO4 x (2 moles Na+ / 1 mole Na2SO4) x 1.20 L = 0.4392 moles of Na+ ions
So, there are approximately 0.4392 moles of Na+ ions in 1.20 liters of the Na2SO4 solution.
To calculate the number of moles of Na+ ion in the given Na2SO4 solution, we need to use stoichiometry.
First, let's write down the balanced chemical equation for the dissociation of Na2SO4:
Na2SO4 -> 2Na+ + SO4²-
From the equation, we can see that for every molecule of Na2SO4, 2 moles of Na+ ions are produced.
Now, let's use the given concentration of the Na2SO4 solution (0.183 M) to calculate the moles of Na2SO4:
Moles of Na2SO4 = Concentration of Na2SO4 in Molarity x Volume of solution in liters
= 0.183 M x 1.20 L
= 0.2196 mol Na2SO4
Since the stoichiometry of the balanced equation tells us that for every one mole of Na2SO4, 2 moles of Na+ ions are produced, we can calculate the moles of Na+ ions:
Moles of Na+ ions = Moles of Na2SO4 x (2 moles Na+ ions / 1 mole Na2SO4)
= 0.2196 mol x (2 mol / 1 mol)
= 0.4392 mol
Therefore, the number of moles of Na+ ion present in 1.20 L of 0.183 M Na2SO4 solution is 0.4392 mol.