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ioda
3 years ago
7

Sodium sulfate dissolves as follows: Na2SO4(s) → 2Na+(aq) + SO42- (aq). How many moles of Na2SO4 are required to make 1.0 L of s

olution in which the Na concentration is 0.10 M?
Chemistry
1 answer:
maxonik [38]3 years ago
8 0

<u>Answer:</u> The number of moles of Na_2SO_4 is 0.05 moles.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

We are given:

Molarity of solution = 0.10 mol/L

Volume of solution = 1 L

Putting values in above equation, we get:

0.10mol/L=\frac{\text{Moles of sodium}}{1.0L}\\\\\text{Moles of sodium}=0.10mol

The chemical reaction for the ionization of sodium sulfate follows the equation:

Na_2SO_4(s)\rightarrow 2Na^+(aq.)+SO_4^{2-}(aq.)

By Stoichiometry of the reaction:

2 moles of sodium ions are produced by 1 mole of sodium sulfate

So, 0.10 moles of sodium ions will be produced by = \frac{1}{2}\times 0.1=0.05moles of sodium sulfate.

Hence, the number of moles of Na_2SO_4 is 0.05 moles.

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