Calculate the molality of a 6.55 kg sample of a solution of the solute CH²Cl²dissolve in the solvent acetone C³H0O if the sample contains 876 g of methylene chloride
Moles of methylene chloride = "\\dfrac{mass}{MM}=\\dfrac{876g}{84.93g\/mol}" =10.314mol
Volume of solution = \dfrac{mass of mixture}{density of acetone}
="\\dfrac{6550g}{0.784g\/cm^3} = 8354.59cm^3"
Molarity og solution;
10.314 mole "\\to" 8354.59cm3
? "\\to" 1000cm3
="\\dfrac{10.314mol x 1000cm^3}{8354.59cm^3} = 1.23M"
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