Answer to Question #170733 in Organic Chemistry for Ash Nash

Question

A solution prepared by dissolving 180.0  mg of a sugar (a molecular compound and a nonelectrolyte) in 1.00 g of water froze at -1.86°C. What is the molar mass of this sugar? The value of K_f is 1.86°C/m.

g/mol?

Answer 1

m_s = 180mg = 0.18g

m_w = 1.00g = 10^-³kg

∆T_f = -1.86

molality × k_f = ∆T_f

molality = -1.86/1.86 = 1mol/kg

molality (m) = number of moles of solute/ mass of solvent =

1 = no. of moles/10-³kg

no. of moles = 10^-³ moles

If 10^-³ moles = 0.18g

1 mole = 180g/mol