The mass of sodium sulfate required to form 250mL of a solution in which the concentration of sodium ion is 0.183M is 6.53 grams.
The mass of a substance can be calculated by multiplying the number of moles in the substance by its molar mass as follows:
Mass = no of moles × molar mass
However, according to this question, sodium sulfate is has a solution with volume 250mL with a concentration of sodium ions of 0.183M. The no of moles must first be calculated as follows:
moles = 0.183 × 0.250 = 0.046moles
Mass of sodium sulfate = 0.046 × 142.04 g/mol = 6.53 grams.
Therefore, 6.53 grams is the mass of the sodium sulfate solution.
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The mass of sodium sulfate required to form 250 mL of solution in which the concentration of sodium ions is 0.183 would be 3.25 grams.
Sodium sulfate is made up of sodium and sulfate ions according to the following equation:
[tex]Na_2SO_4 --- > 2Na^+ + SO_4^2^-[/tex]
From the equation above, a mole of sodium sulfate is made from 2 moles of sodium ions and 1 mole of sulfate ions.
Recall that: mole = molarity x volume
The mole of 0.183 M sodium ion in 250 mL os solution would be:
0.250 x 0.183 = 0.04575 mol
The equivalent mole of sodium sulfate would be:
0.04575/2 = 0.022875 mol
Recall that: mass = mole x molar mass
The molar mass of sodium sulfate is 142.04 g/mol
Mass of 0.022875 mol sodium sulfate = 0.022875 x 142.04
= 3.249165 grams
3.249165 grams to 3 significant figures = 3.25 grams.
Thus, the mass of sodium sulfate required would be 3.25 grams.
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