Answer to Question #174640 in Inorganic Chemistry for Cleo

Question #174640

Determine the concentration of a solution that contains 0.818 grams of Na3PO4 dissolved in 428.0 mL of water in molarity, molality, mole fraction.


1
Expert's answer
2021-03-23T05:54:37-0400

Molar Mass of Na3PO4 = 3(22.989769) + 30.973762 + 4(15.999)

= 163.939069 g/mol

1m = 163.939069 g/mol

? = 0.818g

(1 × 0.818) / 163.939069

= 0.004989658 mol

Molar Mass of H2O = 2.01568 + 15.999

= 18.01468 g/mol

1 m = 18.01468

? = 428

= (1× 428)/18.01468

= 23.758401 moles

Molarity = mol of the solute / L of solution

0.0049896587/(0.428+0.818)

= 0.004004525g/L

Mole fraction = mol of solute/ total moles

= 0.0049896587/( 23.758401 + 0.49896587)

= 0.0002099725

Molality = mol of solute/ Kg of solvent

428 L = 0.428 L

= 0.0049896587/0.428

= 0.0116581 g/L


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