If 1710 g of a sodium citrate is dissolved in 2290 g of watersolvent, calculate

Question

If 1710 g of a sodium citrate is dissolved in 2290 g of watersolvent, calculate the magnitude (not the sign) of the boilingpoint elevation of the solution. Consider if the solute is anelectrolyte.

Molar Mass (g/mol) Na3C6H5O7 258.08 H2O 18.015 Solvent Melting Point (oC) 0.00 Boiling Point (oC) 100.00 Kf (oC/m) 1.858 Kb (oC/m) 0.512
Molar Mass (g/mol) Na3C6H5O7 258.08 H2O 18.015 Solvent Melting Point (oC) 0.00 Boiling Point (oC) 100.00 Kf (oC/m) 1.858 Kb (oC/m) 0.512

Explanation / Answer

T = imkb i = 4 because Na3C6H5O7 - > 3Na+ +C5H6O73- m=molality = moles of solute / kg of solution =((1710g/258.08g/mol)/2.290 kg) kb = .512 T = 4 x ((1710g/258.08g/mol)/2.290 kg) x .512 T = 5.93 (boiling point elevation)

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