Question

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

Answer 1

Answer:

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