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Rank these five scenarios from lowest freezing point (1) to highest freezing point (5). each solution was made using 1.0 kg water as the solvent and the indicated number of moles of each solute. 0.5 mol nacl 1.0 mol k3po4 2.0 mol glucose 1.5 mol ethylene glycol 0.8 mol mgcl2

Respuesta :

Hagrid
Given:

Five solvents dissolved in 1.0 kg water:

a) 0.50 mol NaCl
vant hoff factor, i = 2
ΔT = iKfm
      = 2*1.86*0.50mol/kg = 1.86 C

b) 1.0 mol K3PO4, i = 4
ΔT = iKfm
      = 4*1.86*1.0mol/kg = 7.44 C

c) 2.0 mol Glucose, i = 1
ΔT = iKfm
      = 1*1.86*2.0mol/kg = 3.72 C

d) 1.5 mol Ethylene Glycol, i = 1
ΔT = iKfm
      = 1*1.86*1.50mol/kg = 2.79 C

e) 0.8 mol MgCl, i = 2
ΔT = iKfm
      = 2*1.86*0.80mol/kg = 2.976 C

From lowest to highest freezing point: b, c, e, d, a

your correct on that one