What is the concentration (in Molarity) of ions in solution when 2.13 g of Na_2S

Question

What is the concentration (in Molarity) of ions in solution when 2.13 g of Na_2SO_4 is dissolved in enough water to produce 100.0 mL of solution? 0.150 M 0.300 M 0.450 M 0.600 M In the Enthalpy, Entropy and Free Energy experiment, which one of the following would cause the calculated value of the solubility product constant, K_sp to be too small? Allowing the collected sample to cool prior to beginning the titration. Collecting solid lead chloride from the aqueous solution in addition to the aqueous sample that is then titrated. Titrating until you see red precipitate form. Using a cool pipet, which causes lead chloride to precipitate and not be delivered into the sample to be titrated.

Explanation / Answer

Mass of Na2SO4 = 2.13 g ,

Molar mass of Na2SO4 = 142.04 g/mol

Mole of Na2SO4 = 2.13 g / ( 142.04 g/mol) = 0.015 mol

Volume of solution = 100 ml = 0.1 L

We know Molarity = Moles / volume in Litre

Molarity = 0.015 mol / 0.1 L =0.0015 mol/L = 0.0015M

Since Na2SO4 dissolves in water so it will dissociate into 2 mol of Na+ and 1.0 Mole of SO42-

Concentration of Na+ = 2* 0.0015 M =0.003 M

Concentration of SO42- = 1*0.0015 M = 0.0015 M

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