Concentration and Units - AP Chemistry

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Question

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Answer

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 =

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Question

What is the concentration of Ca in a solution of 1 mol CaCl2 in 1 L of distilled water? (M = molarity, m= molality)

Answer

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.

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Question

What is the concentration if 40 g NaOH is dissolved in 1000 g of water

Answer

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

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Question

Which of the following is equivalent to molarity?

Answer

Molarity, molality, and normality are the three principle ways to measure concentration. Molarity is a measure of moles of solute per liter of solution. Molality is a measure of moles of solute per kilogram of solvent. Normality expressly relates to acids and bases, and is the measure of moles of solute divided by the number of hydrogen equivalents per mole, all divided by liters of solution. Normality is also referred to as "equivalents (of acid) per liter."

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Question

Given that 50 mL of 3M HI was present, what volume would be needed to change the concentration to 0.75M?

Answer

Use the equation M1V1 = M2V2, and plug in the corresponding values to solve for V2:

V2 = (3)(50)/(0.75) = 200mL

We started out 50mL of the solution, so you have to add 150mL to get a final volume of 200mL.

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Question

What is the molality of a solution created by mixing 4.3g $NaCl$ into 43g of water?

Answer

Molality can be defined:

$Molality=\hspace{1 mm}\frac{mol\hspace{1 mm}solute}{kg\hspace{1 mm}solvent}$

It is slightly different from Molarity and has different uses.

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

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Question

How much solid NaOH must be dissolved to make 740mL of a 0.32M solution?

Answer

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.

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

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Question

A 100mL solution is composed of 25% ethanol by volume and water. What is the mass of the solution?

$\rho_{ethanol}=0.789\frac{g}{mL}$

Answer

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.

$25\hspace{ 1 mm}mL\hspace{ 1 mm}EtOH\times\frac{0.789\hspace{ 1 mm}g}{1\hspace{ 1 mm}mL}=19.7\hspace{ 1 mm}g\hspace{ 1 mm}EtOH$

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.

$19.7\hspace{ 1 mm}g+75.0\hspace{ 1 mm}g= 94.7\hspace{ 1 mm}g$

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Question

How many liters of 10M HCl are needed to make 4L of 0.4M solution?

Answer

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

We can rewrite this as .

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

$V = \frac{1.6mol}{10M}=0.16L$

We need 0.16 liters of our 10 molar solution.

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Question

Find the mass of in 10L of water if it is a 2m solution.

Answer

Molality is grams of solute per kilogram of solvent.

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

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 per kilogram of water.

$10L\frac{1kg}{1L}=10kg$

has a molecular weight of $\small 56\frac{g}{mol }$ .

$20mol\frac{56g}{1mol}=1120g\ KOH$

This gives us of .

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Question

Which solution has the highest molarity?

Answer

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.

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

$NaCl:\ \frac{174g\frac{1mol}{58.5g}}{2.0L}=1.49M$

$MgSO_4:\ \frac{240g\frac{1mol}{120.3g}}{3.0L}=0.67M$

$NaNO_3:\ \frac{85g\frac{1mol}{85g}}{2.0L}=0.50M$

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

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Question

A chemist has a bottle containing a 2M aqueous solution of hydrochloric acid. He needs to create a 50mL solution of hydrochloric acid that has a concentration of 0.5M. What is the volume of 2M hydrochloric acid that he should dilute in order to achieve the desired concentration?

Answer

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.

$\small M_{1}v_{1} = M_{2}v_{2}$

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

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

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

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.

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Question

Which of the following aqueous solutions is the most concentrated?

$\small \rho_{water}=1 \frac{g}{mL}$

Answer

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.

$1M\ HCl=\frac{1mol\ HCl}{1L}$

$1m\ HCl=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

$36.5g\ \text{HCl in}\ 1kg\ \text{water}=\frac{1mol\ HCl}{1kg\ water}=\frac{1mol\ HCl}{1L\ water}$

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.

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Question

A chemist wants to turn a 50.0mL solution of $0.60 M\ HNO_{3}$ into a solution. How much water should she add?

Answer

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

Recall:

$M=\frac{mol}{L}$

Plug in known values and solve.

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.

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Question

In order to dilute a 1mL solution that is 0.01M so that the solution is diluted to $5*10^{-3} M,$ , how many milliliters does this solution need to be diluted to?

Answer

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{M_{2}}=V_{2}$

Plugging in the values gives:

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

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Question

What concentration would you have prepared if you diluted 30mL of a 0.350M salt solution to 50mL?

Answer

Use the dilution formula:

$M_{1}V_{1}=M_{2}V_{2}$

Rearranging this equation gives:

$\frac{M_{1}V_{1}}{V_{2}}=M_{2}$

Plugging in the values gives:

$M_{2}=\frac{0.35\ M * 30\ mL}{50\ mL}= 0.21 M$

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

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Question

How many moles are in a 0.010L solution with a concentration that is $5*10^{-3}M$ ?

Answer

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

$\frac{510^{-3}moles}{L} 0.010L = 510^{-5}moles$

Therefore, there are $510^{-5}moles$ in of a $5*10^{-3}M$ solution.

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Question

In order to prepare a needed solution for an experiment, 0.082 grams of was dissolved in water to give a 35mL solution. What is the molarity of this solution?

Answer

In order to calculate the concentration, we must use molarity formula:

$Molarity = \frac{moles\ of\ solute}{volume\ of\ solution\ in\ liters}$

We must use the molecular weight of sodium chloride to calculate the moles of solute:

$MW=(atomic\ weight\ of\ Na)+(atomic\ weight\ of\ Cl)$

$MW_{NaCl}=(23 \frac{g}{mole})+(35 \frac{g}{mole})= 58 \frac{g}{mole}$

$0.082g*\frac{mole}{58g}=0.0014\ moles\ NaCl$

$\frac{0.0014\ moles\ CH_{3}CH_{2}OH}{0.035\L} = 0.040\ M$

Therefore, the concentration in molarity of this solution is 0.040M.

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Question

How many moles are in 1000mL solution with a concentration that is $3*10^{-5}M$ ?

Answer

By simply using the concentration as a conversion factor, the number of moles can be calculated by multiplying the concentration by the number of liters. Before calculating the number of moles, the number of milliliters must be converted to liters using the fact that $1\ liter = 1000\ mL$ .

$\frac{310^{-5}\ moles}{L} 1\ L = 310^{-5}\ moles$

Therefore, there are $310^{-5}\ moles$ in $0.010\ L$ of a $5*10^{-3}M$ solution.

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Question

80.0g NaOH is put into 50000 mL water. What is the molarity of the resulting solution?

Answer

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 =

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