Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605c: 52

59 g

Given: Molality= 1.0 mol/kg and Mass of the solvent=950. g = 0.950 kg Recall: Molality=$\frac{moles\, of\,solute}{mass\, of\, solvent\,in\,kg}$ $\implies$ moles of ethylene glycol= Molality$\times$ mass of solvent in kg $=1.0\,mol/kg\times 0.950\,kg=0.950\,mol$ Mass of ethylene glycol needed = number of moles$\times$ molar mass $=0.950\,mol\times 62.07\,g/mol=59\,g$