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High School Chemistry : High School Chemistry

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Example Questions

Example Question #2 : Concentration And Units

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

Possible Answers:

40 M

4 m

4 M

1 M

1 m

Correct answer: 1 M

Explanation:

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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Example Question #3 : Concentration And Units

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

Possible Answers:

1.6 M

0.00004 M

0.0016 M

0.625 M

0.04 M

Correct answer:

0.04 M

Explanation:

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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Example Question #4 : Concentration And Units

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

Possible Answers:

0.16

8.0

1.6

0.8

0.4

Correct answer:

0.16

Explanation:

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 (molarity)(volume) = moles .

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

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

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

We need 0.16 liters of our 10 molar solution.

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Example Question #5 : Concentration And Units

Which solution has the highest molarity?

Possible Answers:

$\240g\ \text{of}\ MgSO_4\ \text{in}\ 3.0L\ H_2O$

$\138.0g\ \text{of}\ CH_3CH_2OH\ \text{in}\ 3.0L\ H_2O$

$\174g\ \text{of}\ NaCl\ \text{in}\ 2.0L\ H_2O$

$\85.0g\ \text{of}\ NaNO_3\ \text{in}\ 2.0L\ H_2O$

$\486.5g\ \text{of}\ Pb(NO_3)_2\ \text{in}\ 0.75L\ H_2O$

Correct answer:

$\486.5g\ \text{of}\ Pb(NO_3)_2\ \text{in}\ 0.75L\ H_2O$

Explanation:

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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Example Question #6 : Concentration And Units

Which of the following aqueous solutions is the most concentrated?

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

Possible Answers:

$36.5g\ \text{HCl in}\ 1kg\ \text{water}$

$\small \small 1m HCl$

$\small \small 1M HCl$

All of these solutions have the same concentration.

Correct answer:

$\small \small 1M HCl$

Explanation:

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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Example Question #7 : Concentration And Units

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

Possible Answers:

1.71m

0.10m

Cannot be determined

0.71m

0.03m

Correct answer:

1.71m

Explanation:

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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Example Question #1 : Titrations

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

Possible Answers:

9.47g

4.26g

9.47 * 10²g

The sodium hydroxide will boil off with the water

12.8g

Correct answer:

9.47g

Explanation:

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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Example Question #1 : Titrations

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

Possible Answers:

2240g

112g

560g

224g

1120g

Correct answer:

1120g

Explanation:

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 KOH per kilogram of water.

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

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

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

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

This gives us 1120 g of KOH .

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Example Question #1 : Titrations

10mL of a solution of NaOH of unknown concentration mixed with 34mL of $10^{-4}M$ HCl produced a solution with a pH of 7. What is the concentration of the NaOH solution?

Possible Answers:

0.01M

$3.4\cdot 10^{-4}M$

None of these

0.001M

3.4M

Correct answer:

$3.4\cdot 10^{-4}M$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

N_A is the number of acidic hydrogens on the acid, is the molarity of the acid, is the volume of the acid, is the number of basic hydroxides on the base, is the molarity of the base, is the volume of the base

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=3.4\cdot 10^{-4}M$

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Example Question #472 : High School Chemistry

30mL of a solution of NaOH of unknown concentration mixed with 19mL of $10^{-3}M$ HCl produced a solution with a pH of 7. What is the concentration of the NaOH solution?

Possible Answers:

0.0001

None of these

$6.3\cdot 10^{-4}$

$6.3\cdot 10^{-3}$

0.002

Correct answer:

$6.3\cdot 10^{-4}$

Explanation:

For this question use the following formula:

N_A M_A V_A = N_B M_B V_B

Rearrange the equation for the molarity of the base:

$\frac{N_A M_A V_A }{N_B V_B}= M_B$

Plug in known values and solve.

$M_B=6.3\cdot10^{-4}M$

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High School Chemistry : High School Chemistry

Study concepts, example questions & explanations for High School Chemistry

All High School Chemistry Resources

Example Questions

Example Question #2 : Concentration And Units

Example Question #3 : Concentration And Units

Example Question #4 : Concentration And Units

Example Question #5 : Concentration And Units

Example Question #6 : Concentration And Units

Example Question #7 : Concentration And Units

Example Question #1 : Titrations

Example Question #1 : Titrations

Example Question #1 : Titrations

Example Question #472 : High School Chemistry

All High School Chemistry Resources

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