*Consider a 0.15 m strontium chloride aqueous solution.

a. Determine the molality (in m) of all solutes in the solution. Assume that strontium chloride completely ionizes when dissolved in water. All solutes means all cations and anions. The molality that you determine here is the value for m solute. You must explain your answer.

SrCl2 => Sr^2+ + 2Cl^-

m = mols solute/kg solvent
If SrCl2 is 0.15 m then
Sr^2+ must be 0.15 m and Cl^- must be 0.30 m

What is your formula for this?

To determine the molality (m) of all solutes in the 0.15 m strontium chloride (SrCl2) aqueous solution, we first need to understand the concept of molality.

Molality is defined as the amount of substance (moles) of solute dissolved in 1 kilogram of solvent. It is denoted by "m" and expressed in units of mol/kg.

In this case, since strontium chloride completely ionizes when dissolved in water, we need to consider both the strontium ions (Sr2+) and chloride ions (Cl-) as solutes.

The formula for molality is:

molality (m) = moles of solute / mass of solvent (in kg)

Let's calculate the molality of both the strontium ions and chloride ions separately:

1. Strontium ions (Sr2+):
Since 1 mole of SrCl2 dissociates into 1 mole of Sr2+, we can determine the amount of Sr2+ ions by multiplying the initial concentration of SrCl2 by the number of moles of Sr2+ it produces:
moles of Sr2+ = 0.15 mol SrCl2 x 1 mol Sr2+ / 1 mol SrCl2 = 0.15 mol Sr2+

Next, we need the mass of solvent in kilograms. Let's assume we have 1 kg of water as the solvent.

Now we can calculate the molality of Sr2+:
molality of Sr2+ (mSr2+) = 0.15 mol Sr2+ / 1 kg = 0.15 m

2. Chloride ions (Cl-):
Similarly, since 1 mole of SrCl2 dissociates into 2 moles of Cl-, we can determine the amount of Cl- ions by multiplying the initial concentration of SrCl2 by the number of moles of Cl- it produces:
moles of Cl- = 0.15 mol SrCl2 x 2 mol Cl- / 1 mol SrCl2 = 0.30 mol Cl-

Again, considering 1 kg of water as the solvent, we can calculate the molality of Cl-:
molality of Cl- (mCl-) = 0.30 mol Cl- / 1 kg = 0.30 m

Therefore, the molality of all solutes in the 0.15 m strontium chloride aqueous solution is:
molality of Sr2+ (mSr2+) = 0.15 m
molality of Cl- (mCl-) = 0.30 m

To determine the molality (m) of the solutes in the strontium chloride (SrCl2) solution, we need to know the amount of solute (SrCl2) and the mass of the solvent (water).

The molality (m) of a solution is defined as the moles of solute per kilogram of solvent. It can be calculated using the formula:

m = (moles of solute) / (mass of solvent in kg)

In this case, we are given that the concentration of the strontium chloride solution is 0.15 m, which means that there are 0.15 moles of SrCl2 dissolved in 1 kg of water.

First, we need to calculate the moles of SrCl2 in the solution. Since it is mentioned that strontium chloride completely ionizes in water, it means that one mole of SrCl2 will dissociate into one mole of strontium ions (Sr2+) and two moles of chloride ions (Cl-).

Therefore, the moles of solute (SrCl2) in the solution will be the same as the moles of ions (Sr2+ and Cl-).

moles of SrCl2 = moles of Sr2+ = moles of Cl- = 0.15 moles

Next, we need to calculate the mass of water in kg. The mass of water can be calculated using its density, which is approximately 1 g/cm³ or 1000 kg/m³.

mass of water = volume of water x density of water
= 1 L x 1000 g/L / 1000 g/kg
= 1 kg

Now, we can calculate the molality (m) of the solutes:

m = (moles of SrCl2) / (mass of water)
= 0.15 moles / 1 kg
= 0.15 m

Therefore, the molality (m) of all solutes (both cations and anions) in the strontium chloride solution is 0.15 m.