A solution contains 30 wt% water, 25 wt% ethanol, 15 wt% methanol, 12 wt% glycerol, 10 wt% acetic acid, and 8 wt% benzaldehyde. What is the mole fraction of each component?
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- When pure furfural (solvent) is added to a mixture containing 0.2 mass fraction diphenylhexane and 0.8 mass fraction docosane, two separate layers can be obtained. If the composition of the two layers are as shown in the table, how much solvent was added to 500 kg of solution of docosane and diphenylhexane?What will be the new boilng point of 498ml benzene (Kb = 2.53 oC•Kg/mol; Bp = 80.1oC) when mixed with 18g of naphthalene (MW = 128.17 g/mol) to make a non- eletrolyte solution? Density of benzene = 0.876 g/ml)In class, we derived the expression:ln XA =(ΔvH/R) ((1/T)-(1/T°))where A represents the solvent (and B would represent the solute, such that XA + XB = 1).Using the approximations given below, derive the expression:∆Tb = XB (RT°2/ΔvH)where ∆Tb is the change in boiling point between the solution and the pure liquid, T − T°.Approximations:ln(1 − X) ≈ −XTT° ≈ T°2Note: this derived expression is often simplified in General Chemistry as ∆Tb = Kbm where Kb is the boiling point elevation constant, and m is the molality.
- The weight fraction of methanol in an aqueous solution is 0.64. The mole fraction of methanol (x) satisfies a. X < 0.5 b. 0.5 < XM < 0.64 c. xu = 0.5 d. Xu ~ 0.5What will be the new boilng point of 491ml benzene (Kb = 2.53 oC•Kg/mol; Bp = 80.1oC) when mixed with 11g of naphthalene (MW = 128.17 g/mol) to make a non- eletrolyte solution? Density of benzene = 0.876 g/ml) Round you answer in 2 decimal places, unit is not required.1. How many microliters of a 25 \mu M MgCl2.6H2O solution would be needed to provide 152.48 ng of MgCl2.6H2O for a chemical reaction? [ Mwt MgCl2.6H2O = 203.3 g/mole
- Calculate molar concentration of a solution that is 50.0% NaOHby weight and has a specific gravity of 1.52 (FW NaOH= 40.00g/mol)The partial molar volumes of propanone and trichloromethane in a mixture in which the mole fraction of CHCI3 is 0.4693 are 74.166 cm3 mol-1, and 80.235 cm3 mol-1, respectively. What is the volume of a solution of total mass 1.000 kg?The partial molar volumes of acetone and chloroform, in a solution in which the molar fraction of chloroform is 0.4693, are 74.166 cm3mol-1 and 80.235 cm3 mol-1, respectively. What is the volume of 1,000 kg of this solution? Data: M.M. (acetone) = 58.08 gmol-1; M.M. (chloroform) = 119.07 gmol-1.
- 1.Define binary mixture 2. Define density 3. Give an abstract of determination of excess molar volume of binary mixtures of water and methanol by measuring densityWhat is the mole fraction of 35% aqueous ferric hydroxide solution? Fe=56, O=16, H=1 Please take note: 1% = 1g MW solute: ___g/mol MW solvent: ___g /mol mole solute: ___mole mole solvent: ___mole mole fraction solvent:___ mole fraction solute: ___A solution is prepared by dissolving iodine, I2, in carbon tetrachloride, CCl4, at 25 °C. What is the mole fractions of the solute for a solute of molality 0.100 mol kg−1? R = 8.3145 J/Kmol. Mole fraction of solute = 0.015 Mole fraction of solvent = 0.985 (c) change of chemical potential (solvent) = __________ J/mol. 3 sig. fig.