{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Chp2_Various - A 0.227-M solution of X(Mm = 195 g/mol has a...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
A 0.227-M solution of X (M m = 195 g/mol) has a density of 0.959 g/ml. What is the molality of X in the solution? Given Concentration Unit (GCU): 0.227 mol X / 1 L of soln Desired Concentration Unit (DCU): mol X / kg of solvent Note: we have mol X in both, so problem is effectively that of converting 1 L solution to mass of solvent. 1. Since GCU is expressed per L of solution, a convenient “size” sample is 1.0 L. Thus, 0.227 mol X. 2. The only information we have about the solution is the density, allowing us to convert volume of soln to mass soln. 1L * (0.959g/ml)*(1000mL/1L) = 959 g or 0.959 kg of solution. 3. mass of solution = 0.959 kg = mass of solvent + mass of solute . Since we need kg of solvent in the DCU, we must obtain mass of solute. 4. The only information given about the solute is the molar mass. As we have mol X, we convert: 0.227 mol X * (195 g/mol) * (1kg/1000g) = 0.0443kg solute. 5. Mass of solvent = 0.959 – 0.0443 = 0.915 kg. 6. Molality = 0.227 mol X/ 0.915 kg solvent = 0.248 m. Colligative Properties: P. Exp 2.14, p. 53. 62.3 g CaCl2 in 100 mL of water at 20ºC.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}