Textbook Question
Calculate the freezing point and boiling point of each aqueous solution, assuming complete dissociation of the solute. a. 0.100 m K2S b. 21.5 g of CuCl2 in 4.50⨉102 g water
1430
views
Ch.13 - Solutions
Chapter 13, Problem 1
Examine the structures of phenol and naphthol shown here and explain why phenol is so much more soluble in water at 25 °C (87 g/L) compared to naphthol (0.74 g/L).
Verified step by step guidance1
<Phenol (C6H5OH) and naphthol (C10H7OH) are both aromatic compounds with a hydroxyl group (-OH) attached to their benzene rings.>
<The solubility of a compound in water is influenced by its ability to form hydrogen bonds with water molecules.>
<Phenol has a single benzene ring, which allows the hydroxyl group to interact more effectively with water, forming hydrogen bonds.>
<Naphthol, on the other hand, has two fused benzene rings, which makes the molecule bulkier and reduces the accessibility of the hydroxyl group to water molecules, thus limiting hydrogen bonding.>
<The increased hydrogen bonding capability of phenol compared to naphthol results in its higher solubility in water.>
Related Practice
Textbook Question
Pick an appropriate solvent from Table 13.3 to dissolve each substance. State the kind of intermolecular forces that would occur between the solute and solvent in each case. a. isopropyl alcohol (polar, contains an OH group) c. vegetable oil (nonpolar) d. sodium nitrate (ionic)
Open Question
Use Henry’s Law to determine the molar solubility of helium at a pressure of 1.0 atm and 25 °C.
Open Question
You can purchase hydrochloric acid in a concentrated form that is 37.0% HCl by mass and that has a density of 1.20 g>mL. Describe exactly how to prepare 2.85 L of 0.500 M HCl from the concentrated solution.
Open Question
Calculate the freezing point and boiling point of a solution containing 7.55 g of ethylene glycol (C2H6O2) in 85.7 mL of ethanol. Ethanol has a density of 0.789 g/cm³.