1. Sugar can be dissolved in water because of the presence of bonds between _______ in its molecule.

2. If the maximum amount of a compound that can dissolve in 100 mL of water is 22 g, then a solution of 20 g of that substance in 100 mL of water is said to be _______.

3. A solution of 25 g of the substance described in question 2 in 100 mL of water would be called ________.

4. In most cases, one way to increase the solubility of a solid substance in water is to increase the _______ of the solution.

5. A more precise way for describing the concentration of a solution than to describe it as concentrated or dilute is to state its ___________.

6. To make 5 L of 1M aluminum nitrate (Al(NO3)3), you would have to dissolve ________ of the solute in 5.0 L of solution.

7. 400 mL of a 0.1M barium nitrate (Ba(NO3)2) solution consists of _______ of solute dissolved in 400 mL of solution.

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1. Sugar can be dissolved in water because of the presence of bonds between the atoms in its molecule. The main bonds responsible for sugar's solubility in water are the hydrogen bonds formed between the oxygen and hydrogen atoms in water molecules and the hydroxyl groups (-OH) in sugar molecules.

To get the answer to this question, you can look up the chemical structure of sugar (sucrose) and analyze its molecular bonds.

2. If the maximum amount of a compound that can dissolve in 100 mL of water is 22 g, then a solution of 20 g of that substance in 100 mL of water is said to be saturated.

To determine the solubility of a compound, you can perform a saturation experiment, gradually adding the compound to a fixed volume of water until no more dissolves. The mass of the compound that dissolves at saturation represents its solubility in that specific volume of solvent.

3. A solution of 25 g of the substance described in question 2 (with a maximum solubility of 22 g in 100 mL of water) in 100 mL of water would be called supersaturated.

Supersaturation occurs when more solute is dissolved in a solvent than can typically be dissolved under normal conditions. In this case, the solution would be unstable and prone to precipitation.

4. In most cases, one way to increase the solubility of a solid substance in water is to increase the temperature of the solution.

For many solid solutes, an increase in temperature leads to an increase in kinetic energy of the solvent particles, enhancing their ability to break the attractions holding the solute particles together. This results in a greater solute-solvent interaction and increased solubility.

5. A more precise way of describing the concentration of a solution than to state it as concentrated or dilute is to state its molarity.

Molarity (M) is a unit of concentration that represents the number of moles of solute dissolved in one liter (L) of solution. It is calculated by dividing the amount of solute (in moles) by the volume of solution (in liters).

6. To make 5 L of 1M aluminum nitrate (Al(NO3)3), you would have to dissolve 53.3 g of the solute in 5.0 L of solution.

To calculate the amount of solute required for a specific molar concentration, you need to know the molar mass of the compound. In this case, the molar mass of Al(NO3)3 is 213 g/mol. Multiply the desired molarity (1M) by the desired volume (5 L) and the molar mass to obtain the mass of the solute required.

7. 400 mL of a 0.1M barium nitrate (Ba(NO3)2) solution consists of 1.6 g of solute dissolved in 400 mL of solution.

To calculate the amount of solute in a solution given its molarity, you need to know the molar mass of the compound. In this case, the molar mass of Ba(NO3)2 is 261.34 g/mol. Multiply the molarity (0.1M) by the volume of solution (400 mL) and the molar mass to find the mass of the solute. Convert the volume to liters (0.4 L) before the calculation.