The definition of molality is moles of solute in 1 kg of the solvent, whereas molarity is the number of moles of solute per 1 L of solutioin. Since 1 mol of CaCl2 is added to 1 L of water, this means that the volume of the final solution is greater than 1 L. Thus, molality is the more accurate concentration determinant, since the solution is probably close to 1 L. 

In order to answer this question, it helps to know that 1 kilogram of water is equal to 1 liter of water, due to its density. Two of the above options refer to a 1m solution of hydrochloric acid. The other is a 1M solution. 

\(\displaystyle 1M\ HCl=\frac{1mol\ HCl}{1L}\)

\(\displaystyle 1m\ HCl=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}\)

\(\displaystyle 36.5g\ \text{HCl in}\ 1kg\ \text{water}=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}\)

All three of the options have the same amount of hydrochloric acid (one mole). For molarity, the hydrochloric acid is diluted with water until one liter of solution is created. For molality, one mole of HCl is added to one kilogram of water. Since one kilogram of water is one liter, this becomes the same concentration.

One a very small level, the 1M HCl solution will be slightly more concentrated. Creating a molal solution does not take into account the volume of the solute. If, for example, 100 cubic centimeters of HCl were added to one kilogram of water, the resulting volume would be more than one liter, making the concentration slightly less than 1M. This discrepancy is usually not accounted for in basic chemistry, but you should be familiar with the concept.

Molality can be defined:

\(\displaystyle Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}\)

It is slightly different from Molarity and has different uses.

\(\displaystyle Molality=\hspace{1 mm}\frac{4.3\hspace{1 mm}g\hspace{1 mm}NaCl}{43\hspace{1 mm}g\hspace{1 mm}H_20}\times\frac{1000\hspace{1 mm}g\hspace{1 mm}H_2O}{1\hspace{1 mm}kg\hspace{1 mm}H_2O}\times\frac{1\hspace{1 mm}mole\hspace{1 mm}NaCl}{58.44\hspace{1 mm}g\hspace{1 mm}NaCl}=1.71\hspace{1 mm}m\)

This problem can be solved by stoichiometry. Remember that 0.32M gives us the moles of NaOH per liter, and solve for the number of moles per 0.740L.

\(\displaystyle 740\hspace{1 mm}mL\times\frac{1\hspace{1 mm}L}{1000\hspace{1 mm}mL}\times\frac{0.32\hspace{1 mm}moles\hspace{1 mm}NaOH}{1\hspace{1 mm}L}\times\frac{40\hspace{1 mm}g\hspace{1 mm}NaOH}{1\hspace{1 mm}mole\hspace{1 mm}NaOH}=9.47\hspace{1 mm}g\hspace{1 mm}NaOH\)

Molality is grams of solute per kilogram of solvent.

\(\displaystyle m=\frac{mol}{kg}\rightarrow mol=(m)(kg)\)

Water has a density of one gram per mililiter, so one liter of water equal to one kilogram. If we have a 2m solution, that means we have two moles of \(\displaystyle KOH\) per kilogram of water.

\(\displaystyle 10L*\frac{1kg}{1L}=10kg\)

\(\displaystyle mol=(m)(kg)=(2m)(10kg)=20mol\)

\(\displaystyle KOH\) has a molecular weight of \(\displaystyle \small 56\frac{g}{mol }\).

\(\displaystyle 20mol*\frac{56g}{1mol}=1120g\ KOH\)

This gives us \(\displaystyle 1120 g\) of \(\displaystyle KOH\).

In order to dilute the concentrated acid, we need to find the amount of concentrated acid that will be diluted to 50mL of total solution. We can find the volume of concentrated acid necessary by setting the final volume and concentration equal to the initial concentration and unknown volume.

\(\displaystyle \small M_{1}v_{1} = M_{2}v_{2}\)

The initial concentration is 2M, the final concentration is 0.5M, and the final volume is 50mL

\(\displaystyle \small (2M)v_{1} = (0.5M)(50 mL)\)

\(\displaystyle \small v_{1} = \frac{(0.5M)(50mL)}{2M}=12.5 mL\)

This means that 12.5mL of concentrated acid needs to be diluted to 50mL of solution. This will result in a solution with a concentration of 0.5M.

To solve this problem, we may use the following equation relating the molarity and volume of two solutions:

\(\displaystyle M_1V_1=M_2V_2\)

Recall:

\(\displaystyle M=\frac{mol}{L}\)

Plug in known values and solve.

\(\displaystyle (0.6M)(50mL)=(0.4M)(V_2)\)

\(\displaystyle V_2=75mL\)

However, this is not the final answer. The whole volume of the second, 0.4M solution is 85mL. Thus the chemist needs to add 25mL of water to the original solution to obtain the desired concentration.

Use the dilution formula:

\(\displaystyle M_{1}V_{1}=M_{2}V_{2}\)

Rearranging this equation gives:

\(\displaystyle \frac{M_{1}V_{1}}{M_{2}}=V_{2}\)

Plugging in the values gives:

\(\displaystyle V_{2}=\frac{0.01\ M * 1\ mL}{5*10^{-3}\ M}= 2\ mL\)

Use the dilution formula:

\(\displaystyle M_{1}V_{1}=M_{2}V_{2}\)

Rearranging this equation gives:

\(\displaystyle \frac{M_{1}V_{1}}{V_{2}}=M_{2}\)

Plugging in the values gives:

\(\displaystyle M_{2}=\frac{0.35\ M * 30\ mL}{50\ mL}= 0.21 M\)

Therefore, after diluting the solution to 50mL, the solution concentration would be lowered from 0.35M to 0.21M.

By using the concentration as a conversion factor, the number of moles can calculated by multiplying the concentration by the number of liters.

\(\displaystyle \frac{5*10^{-3}moles}{L} * 0.010L = 5*10^{-5}moles\)

Therefore, there are \(\displaystyle 5*10^{-5}moles\) in \(\displaystyle 0.010 L\) of a \(\displaystyle 5*10^{-3}M\) solution.