What is w/v% solution of fe(oh)3 (30% dissociation of 50% dissolved amount) will have boiling point of 223℉
Answers
Answer:
Explanation:
The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition.
\text{mol\;C}_2\text{H}_4(\text{OH})_2 = 2220\;\text{g}\;\times\;\frac{1\;\text{mol\;C}_2\text{H}_4(\text{OH})_2}{62.07\;\text{g\;C}_2\text{H}_4(\text{OH})_2} = 35.8\;\text{mol\;C}_2\text{H}_4(\text{OH})_2
\text{mol\;H}_2\text{O} = 2000\;\text{g}\;\times\;\frac{1\;\text{mol\;H}_2\text{O}}{18.02\;\text{g\;H}_2\text{O}} = 11.1\;\text{mol\;H}_2\text{O}
X_{\text{ethylene\;glycol}} = \frac{35.8\;\text{mol\;C}_2\text{H}_4(\text{OH})_2}{(35.8\;+\;11.1)\;\text{mol\;total}} = 0.763
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
(b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute:
2220\;\text{g\;C}_2\text{H}_4(\text{OH})_2\;(\frac{\text{mol\;C}_2\text{H}_2(\text{OH})_2}{62.07\;\text{g}}) = 35.8\;\text{mol\;C}_2\text{H}_4(\text{OH})_2
Then, convert the mass of the water from grams to kilograms:
2000\;\text{g\;H}_2\text{O}\;(\frac{1\;\text{kg}}{1000\;\text{g}}) = 2\;\text{kg\;H}_2\text{O}
Finally, calculate molarity per its definition:
\begin{array}{r @{{}={}} l} \text{molality} & \frac{\text{mol\;solute}}{\text{kg\;solvent}} \\[1em] \text{molality} & \frac{35.8\;\text{mol\;C}_2\text{H}_4(\text{OH})_2}{2\;\text{kg\;H}_2\text{O}} \\[1em] \text{molality} & 17.9\;m \end{array}
Check Your Learning
What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH3, dissolved in 125 g of water?
Answer:
7.14 × 10−3; 0.399 m
Example 2
Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Solution
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as:
\frac{3.0\;\text{mol\;NaCl}}{1.0\;\text{kg\;H}_2\text{O}}
The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg
1.0\;\text{kg\;H}_2\text{O}\;(\frac{1000\;\text{g}}{1\;\text{kg}})(\frac{\text{mol\;H}_2\text{O}}{18.02\;\text{g}}) = 55\;\text{mol\;H}_2\text{O}
and then substituting these molar amounts into the definition for mole fraction.
\begin{array}{r @{{}={}} l} X_{\text{H}_2\text{O}} & \frac{\text{mol\;H}_2\text{O}}{\text{mol\;NaCl}\;+\;\text{mol\;H}_2\text{O}} \\[0.75em] X_{\text{H}_2\text{O}} & \frac{55\;\text{mol\;H}_2\text{O}}{3.0\;\text{mol\;NaCl}\;+\;55\;\text{mol\;H}_2\text{O}} \\[0.75em] X_{\text{H}_2\text{O}} & 0.95 \\[0.75em] X_{\text{NaCl}} & \frac{\text{mol\;NaCl}}{\text{mol\;NaCl}\;+\;\text{mol\;H}_2\text{O}} \\[0.75em] X_{\text{NaCl}} & \frac{3.0\;\text{mol\;NaCl}}{3.0\;\text{mol\;NaCl}\;+\;55\;\text{mol\;H}_2\text{O}} \\[0.75em] X_{\text{NaCl}} & 0.052 \end{array}