Question

Determine the molality of a solution made by dissolving 0.134 moles of glucose in 471 mL of water. Assume the density of water is 1.00 g/mL.

Answer 1

Solution cu'. Given that is moles of Gulcose = 0:134 mole. volume of water = 471m Density of water = 1.ooglmi so mass of water = 471 me x govog my B = 4719 or 0.471kg we know that: i memass) = VXD Vivolume) OlDensity) ... 1kg = 10oog molality - moles of solute 0 kg of solvent Put the value in above formula mmolality)= 0.134 mole 0:411 kg m = 0.285 mol/kg Ango Que 2 Given that is mass of CH₂OH= 28.19 molar mass of CH3OH=32g/mol Soi mole of CH3OH= 28.19 >> 0.878 mole. 32g/mol mass of water = 5139 or 0.513 kg So molality (m)= 0.878mole 0:513 kg m= 1.71 mol/kg Ansi