First we determine the mass of the ethanol in solution using its density. Using the percent by volume of ethanol, we know that there are 25mL of ethanol in a 100mL solution. The remaining 75mL are water.

\(\displaystyle 25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH\)

Since the density of water is 1g/mL, we know that the mass of 75mL of water is 75g. The total mass is the sum of the ethanol and the water.

\(\displaystyle 19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g\)

first of all, M = molar; m = molal- M = mol solute/ L of solution; m = mol solute/ kg solvent

you have 40 g NaOH * 1 mol/40 g = 1 mol

1000 g of water is equivalent to 1 L

1 mol/L = 1M

Molarity = mol solute / L soution

mol solute = 80 g NaOH * 1 mol / 40 g = 2 mol

L solution = 50000 mL water * 1 L/1000 mL = 50 L

2 mol / 50 L = 

A simple calculation can be done to perform any solution dilution problem. We know our equation \(\displaystyle molarity = \frac{moles}{volume}\).

We can rewrite this as \(\displaystyle (molarity)(volume) = moles\).

Using this formula, we take the old solution and set it equal to the new solution.

\(\displaystyle (10 M)(V) = (0.4M)(4L)=1.6mol\)

\(\displaystyle V = \frac{1.6mol}{10M}=0.16L\)

We need 0.16 liters of our 10 molar solution.

This question requires us to calculate molarity for each answer choice. It is important to add everything correctly and be careful with more complex compounds.

Molarity is simply moles of solute over liters of solution. The correct answer, after trying each, is the answer with lead (II) nitrate, as it gets us a molarity of 2.

\(\displaystyle Pb(NO_3)_2:\ \frac{486.5g*\frac{1mol}{331g}}{0.75L}=1.96M\)

\(\displaystyle NaCl:\ \frac{174g*\frac{1mol}{58.5g}}{2.0L}=1.49M\)

\(\displaystyle MgSO_4:\ \frac{240g*\frac{1mol}{120.3g}}{3.0L}=0.67M\)

\(\displaystyle NaNO_3:\ \frac{85g*\frac{1mol}{85g}}{2.0L}=0.50M\)

\(\displaystyle CH_3CH_2OH:\ \frac{138g*\frac{1mol}{46g}}{3.0L}=1.0M\)

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 \(\displaystyle KOH\).

For this question use the following formula: 

\(\displaystyle N_A M_A V_A = N_B M_B V_B\)

\(\displaystyle N_A\) is the number of acidic hydrogens on the acid, _{\(\displaystyle M_A\)} is the molarity of the acid, _{\(\displaystyle V_A\)} is the volume of the acid, _{\(\displaystyle N_B\)} is the number of basic hydroxides on the base, _{\(\displaystyle M_B\)} is the molarity of the base, _{\(\displaystyle V_B\)} is the volume of the base

Rearrange the equation for the molarity of the base:

\(\displaystyle \frac{N_A M_A V_A }{N_B V_B}= M_B\)

Plug in known values and solve.

\(\displaystyle M_B=3.4\cdot 10^{-4}M\)