We dissolve 10.0 g of each of the following in 800 g of water. In which solution would the molality of solute be highest? 1. All would be the same. 2. sodium iodide 3. potassium iodide 4. rubidium nitrate 5. ammonium nitrate

The highest molality will be for the NH₄NO₃ . Option 5.

Explanation:

In order to determine which of the salts has the highest molality we must know the moles of each

10 g / 149.89 g/mol = 0.0667 moles NaI

10 g / 166 g/mol = 0.0602 moles KI

10 g / 147.4 g/mol = 0.0678 moles of RbNO₃

10 g / 80 g/ mol = 0.125 moles of NH₄NO₃

To determine molality we convert the mass of solvent (water) from g to kg

800 g . 1kg / 1000 g = 0.8 kg

Molality means the moles of solute in 1kg of solvent

0.0667 moles NaI / 0.8 kg of water = 0.083 m

0.0602 moles KI / 0.8 kg of water = 0.0752 m

0.0678 moles of RbNO₃ / 0.8 kg of water = 0.0847 m

0.125 moles of NH₄NO₃ / 0.8 kg of water = 0.156 m

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa