Sodium iodide is given by the formula NaI, and has a molecular weight of 150g/mol. Molarity is found by dividing the moles of solute by liters of solvent.

We find the moles of sodium iodide by using the mass (300g) and molecular weight.

We know our volume is 1L, so now we can solve for the molarity.

To solve this question we must first determine the number of moles of HCl in the initial solution. Converting 100mL into 0.1L, there is initially 0.6mol of HCl in the solution.

When we dilute the solution with 500mL of water, the final volume of solution is 600mL.

To find molarity we can divide number of moles of solute by the total volume of solution.

The final concentration is 1M. 

Molality is the number of moles of solute divided by kilograms of solvent. Because we are given the solute in mmol and the solvent in grams, we must convert them to moles and kilograms, respectively.

300 mmol is 0.3 moles.  Additionally 60 grams is 6 x 10^-2 kg.  

Now we can solve for the molality.

20 grams of ammonia is approximately 1.17 moles of ammonia, so both the molarity and molality of this solution would not approach 2.

Now, we need to determine whether molarity or molality is more concentrated. It helps to remember for the MCAT that 1 liter of water is equal to 1 kilogram of water. Molality uses the denomination of kilograms of water, and molarity mixes the solute until liters of solution are created. Since the mixed solution will incorporate the solute, it will require less than 1 kilogram of water. As a result, 2M is more concentrated than 2m.

First, find the mass of copper (II) sulfate in one liter of solution, using the molar mass of copper (II) sulfate.

We now have the density of the copper (II) sulfate in the solution. Using the total density of the solution, we can calculate the contribution from water to this density.

Now that we know the mass of water per liter of solution, we need to convert to moles.

We now have the moles of copper (II) sulfate per liter and the moles of water per liter, allowing us to find the molar ratio in the solution.

Recall that molality is defined as:

In this problem there is one half mole of sucrose per one kilogram of water.

Our goal is to calculate the mole fraction:

We need to calculate the number of moles of water in one kilogram of solution.

We now know that, in one kilogram of solution, there will be  of sucrose and of water. Use these values in the mole fraction equation to solve for the fraction of sucrose.

Concentration can be measured in several different ways. Molality is moles of solute per kilogram of solvent. Molarity is moles of solute per liter of solution. Normality is equivalents of protons per liter of solution.

In our question, we are given the molality of the solution.

When using molality, it is important to realize that the solute occupies volume. One kilogram of water is equal to one liter of water, but there is an additional volume of four moles of solid sodium chloride that is added to the solution. A one-liter sample of this solution will, thus, contain less than one liter of water.

Use the equation for molarity, , first to figure out the number of moles of solute present.

Solving this gives 0.225 moles of solute. Using the same equation again with the final molarity will give the final volume.

Finally, the amount of solvent added is the difference between initial and final total volumes.

0.643L – 0.300L = 0.342L added

In this particular scenario, molarity can be used interchangeably with normality. The product of the initial normality and initial volume will be equal to the product of the final normality and final volume.

Our initial normality is 0.7N and our initial volume is 0.02L. We also know our final volume is 1L.

Solve for the final concentration.