What is the mass percent of an aqueous sodium hydroxide solution in which the mo

Question

What is the mass percent of an aqueous sodium hydroxide solution in which the mole fraction of NaOH is 0.231? The density of the solution is 1.4339 g/mL. 33.1% 40.0% 6.21% 6.44% 68.9% When a 29.4-g sample of an unknown compound is dissolved in 500. g of benzene, the freezing point of the resulting solution is 3.77 degree C. The freezing point of pure benzene is 5.48 degree C, and K_f for benzene is 5.12 degree C/m. Calculate the molar mass of the unknown compound. 17.2 g/mol 176. g/mol 352 g/mol 88.0 g/mol 151 g/mol The standard enthalpy of fusion of chlorobenzene is 9.55 kJ/mol at its melting point, 227.9 K. What is the standard change in entropy for the melting of chlorobenzene at its melting point? 0.0419 J/(mol middot K) 0.0191 J/(mol middot K) -211 J/(mol middot K) -9.55 J/(mol middot K) 41.9 J/(mol middot K)

Explanation / Answer

Answer:

38) Option b = 40%

Given data:

Mole fraction of NaOH = 0.231

We have only two components, water and NaOH

So, the mole fraction of water is 1 - 0.231 = 0.769

We have 0.231 mol NaOH dissolved in 0.769 mol of water.

Molecular weight of NaOH = 39.99 g/mol

Mass of NaOH = moles * MW = 0.231 moles * 39.99 g/mol = 9.237 g of NaOH

Molecular weight of H2O = 18 g/mol

Mass of Water = moles * MW = 0.769 moles * 18 g/mol = 13.842 g of Water

Total mass = 9.237 + 13.842 = 23.079 g

Therefore, mass percent of NaOH = Individual mass of NaOH / Total mass = 9.237 / 23.079 = 0.4 = 40 %

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