A 16.96 g mixture of sugar (C12H22O11) and table salt (NaCl) is dissolved in 211

Question

A 16.96 g mixture of sugar (C12H22O11) and table salt (NaCl) is dissolved in 211 g of water. The freezing point of the solution was measured as -3.47 °C. Calculate the mass percent of sugar in the mixture. A list of Kf values can be found here.

Attached is above the link for Kf values.

Thanks

Sapling Learning A 16.96 g mixture of sugaz0n) and table salt (NaCI) is dissolved in 211 g of water. The freezing point of the solution was measured as-347 . Calculate the mass percent of sugar in the mixture. A list of values can be found here Number Previous Give Up & View Solution O Check Answer ) Next H Exit Hint

Explanation / Answer

Suppose amount of NaCl = X g

Amount of sugar = (16.96 – X) g

Number of moles of NaCl = Mass/molar mass = (X/58.44)

Number of moles of ions (Na+ and Cl-) = 2 x (X/58.44) = 2X/58.44

Number of moles of sugar = (16.96 – X)/342.30

Number of moles of solutes = 2X/58.44 + (16.96 – X)/342.30 = (684.6 X -58.44 X +991.1424) /20004.012

                                                   = (626.16 X + 991.1424)/20004.012

Molality (m) = Moles of solute/Weight of solvent in KG

                      m = (626.16 X + 991.1424)/(20004.012 x 0.211)

                           = (626.16 X + 991.1424)/4220.002532

Delta Tf = Kf x m

3.74 = 1.86 x (626.16 X + 991.1424)/4220.00

8485.38 = 626.16 X + 991.14

X = 11.97 g

% of NaCl = (11.97/16.96)x100 = 70.58 %

% of Sugar = 100 – 70.58 = 29.42 %

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