Molarity, Molality, Normality - AP Chemistry

Card 0 of 36

Question

A solution is prepared by dissolving $10.2\text{g of glucose,}\text{ C}_6\text{H}_{12}\text{O}_6\text{,}$ in $405\text{g}$ of water. The final volume of the solution is $414\text{mL}$ . Find the concentration of the solution in units of molality.

Answer

Recall how to find the molality of a solution:

$\text{Molality}=\frac{\text{moles of solute}}{\text{kg of solvent}}$

First, start by finding the moles of glucose that we have. The molar mass of glucose is $180.16\text{g/mol}$ .

$10.2\text{g of glucose}\cdot\frac{\text{1 mole glucose}}{180.16\text{g of glucose}}=0.0566\text{ moles of glucose}$

Next, convert the grams of water into kilograms.

$405\text{g}\cdot\frac{1\text{kg}}{1000\text{g}}=0.405\text{kg}$

Now, plug in the moles of glucose and kilograms of water into the equation for molality.

$\text{Molality}=\frac{0.0566}{0.405}=0.140m$

Compare your answer with the correct one above

Question

A solution of hydrogen peroxide is by mass. What is the molarity of the solution? Assume that the solution has a density of $1.01\text{g/mL}$ .

Answer

Start by assuming that we have $1.00\text{L}$ of this solution. Recall that hydrogen peroxide has a molecular formula of $\text{H}_2\text{O}_2$ .

Use the given density to find the mass of the solution.

$1.00\text{L}\cdot\frac{1000\text{mL}}{1\text{L}}\cdot\frac{1.01\text{g}}{\text{mL}}=1010\text{g}$

Next, find the mass of the hydrogen peroxide present in the solution.

$1010\text{g}(0.152)=153.52\text{g}$

Convert the mass of hydrogen peroxide into moles of hydrogen peroxide.

$153.52\text{g of H}_2\text{O}_2\cdot\frac{1\text{ mole H}_2\text{O}_2}{31.04\text{g H}_2\text{O}_2}=4.95\text{ moles of H}_2\text{O}_2$

Recall how to find the molarity of a solution:

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

Since we have $1.00\text{L}$ of solution, the molarity is .

Compare your answer with the correct one above

Question

What is the molarity of a solution in which sodium hydrogen carbonate is dissolved in a solution?

Answer

sodium hydrogen carbonate dissolved in a solution has .

The first step is to calculate how many moles of $NaHCO_{3}$ are present. $211 g NaHCO_{3}\frac{1 mol NaHCO_{3}}{84.006 g NaHCO_{3}}=2.51 mol NaHCO_{3}$

We calculate molarity with the following equation:

$M = \frac{moles of solute}{liters of solution }$

$M = \frac{2.51 moles NaHCO_{3}}{10 L} = 0.251 M$

Compare your answer with the correct one above

Question

How many milliliters of a $5.0 M CuSO_{4}$ solution are needed to prepare of $0.500 M CuSO_{4}$ ?

Answer

of $5.0 M CuSO_{4}$ solution are needed to prepare of $0.500 M CuSO_{4}$ .

We can use the formula $M_{1}V_{1}= M_{2}V_{2}$

Therefore, $(5.00 M_{1})(V_{1})=(0.350 M_{2})(0.500 L)$

$V_{1}= 0.035 L = 35 mL$

Compare your answer with the correct one above

Question

What are the concentrations of aluminum and sulfate in a 3.0 M solution of aluminum sulfate?

Answer

The relative concentrations of aluminum sulfate $(Al_{2}(SO_{4})_{3})$ are

$6.0 M Al^{3+}$

$9.0 M SO_{4}^{2-}$

$\frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{2 mol Al^{3+}}{1 mol Al_{2}(SO_{4})_{3}}=6.0 M Al^{3+}$

$\frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{3 mol SO_{4}^{2-}}{1 mol Al_{2}(SO_{4})_{3}}=9.0 M Al^{3+}$

Compare your answer with the correct one above

Question

What is the molarity of a solution containing of solution containing of ?

Answer

Molarity moles of solute per liter of solution.

We are given that there are of and of solution.

First lets convert the of solution to liters. It is easiest to use scientific notion in all calculations for easy simplification:

=

$\frac{7.510^{2}ml1L}{(10^3ml)}=7.510^{-1}L$

Now we must convert of to moles to get moles of solute per liter of solution.

This is done by dividing by its molecular weight which is . Once again we can use scientific notion to simplify calculations:

$3.510g\ MgCO_31mol\ MgCO_3\frac{1}{(8.41310g\ MgCO_3)}$

Now we can divide by $7.510^{-1}L$ solution to get the molarity.

$4.1610^{-1}mol\ MgCO_3\frac{1}{7.5*10^{-1}L}$

Compare your answer with the correct one above

Question

What is the molality of a solution made by adding of to of water?

Answer

Molality () is defined as moles of solute per kg of solvent.

is the solute (it is what is being dissolved) and water is the solvent (what is doing the dissolving).

Let’s start converting of to moles by dividing by its molecular weight using scientific notion through the entire molality calculations as necessary to simplify calculations.

$9.5g\ NaCl1mol\ NaCl\frac{1}{5.8410g\ NaCl}=0.163mol\ NaCl$

Now we must convert the of water to then divide by it to get the molality:

$310^{2}g\ H_2O1kg\frac{1}{10^3g}=310^{-1}kg\ H_2O$

Molality of solution:

$=1.6310^{-1}mol\ NaCl\frac{1}{310^{-1}kg\ H_2O}$

=

Compare your answer with the correct one above

Question

How many of water are needed to dilute to ?

Answer

In order to solve for the volume of water needed we must use this equation:

solve for since we are looking for the final volume of water needed to dilute the existing solution:

Compare your answer with the correct one above

Question

Which of the following choices is characteristic of molality?

Answer

Molarity, molality, and normality are all units of concentration in chemistry. Molarity () is defined as the number of moles of solute per liter of solution. Molality () is defined as the number of moles of solute per kilogram of solvent. Normality () is defined as the number of equivalents per liter of solution. Molality, as compared to molarity, is also more convenient to use in experiments with significant temperature changes. This is because the volume of a solution increases with temperature, and heating causes molarity to decrease; however, since molality is based on masses rather than volumes, molality remains unchanged.

Compare your answer with the correct one above

Question

A solution is prepared by dissolving $10.2\text{g of glucose,}\text{ C}_6\text{H}_{12}\text{O}_6\text{,}$ in $405\text{g}$ of water. The final volume of the solution is $414\text{mL}$ . Find the concentration of the solution in units of molality.

Answer

Recall how to find the molality of a solution:

$\text{Molality}=\frac{\text{moles of solute}}{\text{kg of solvent}}$

First, start by finding the moles of glucose that we have. The molar mass of glucose is $180.16\text{g/mol}$ .

$10.2\text{g of glucose}\cdot\frac{\text{1 mole glucose}}{180.16\text{g of glucose}}=0.0566\text{ moles of glucose}$

Next, convert the grams of water into kilograms.

$405\text{g}\cdot\frac{1\text{kg}}{1000\text{g}}=0.405\text{kg}$

Now, plug in the moles of glucose and kilograms of water into the equation for molality.

$\text{Molality}=\frac{0.0566}{0.405}=0.140m$

Compare your answer with the correct one above

Question

A solution of hydrogen peroxide is by mass. What is the molarity of the solution? Assume that the solution has a density of $1.01\text{g/mL}$ .

Answer

Start by assuming that we have $1.00\text{L}$ of this solution. Recall that hydrogen peroxide has a molecular formula of $\text{H}_2\text{O}_2$ .

Use the given density to find the mass of the solution.

$1.00\text{L}\cdot\frac{1000\text{mL}}{1\text{L}}\cdot\frac{1.01\text{g}}{\text{mL}}=1010\text{g}$

Next, find the mass of the hydrogen peroxide present in the solution.

$1010\text{g}(0.152)=153.52\text{g}$

Convert the mass of hydrogen peroxide into moles of hydrogen peroxide.

$153.52\text{g of H}_2\text{O}_2\cdot\frac{1\text{ mole H}_2\text{O}_2}{31.04\text{g H}_2\text{O}_2}=4.95\text{ moles of H}_2\text{O}_2$

Recall how to find the molarity of a solution:

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

Since we have $1.00\text{L}$ of solution, the molarity is .

Compare your answer with the correct one above

Question

What is the molarity of a solution in which sodium hydrogen carbonate is dissolved in a solution?

Answer

sodium hydrogen carbonate dissolved in a solution has .

The first step is to calculate how many moles of $NaHCO_{3}$ are present. $211 g NaHCO_{3}\frac{1 mol NaHCO_{3}}{84.006 g NaHCO_{3}}=2.51 mol NaHCO_{3}$

We calculate molarity with the following equation:

$M = \frac{moles of solute}{liters of solution }$

$M = \frac{2.51 moles NaHCO_{3}}{10 L} = 0.251 M$

Compare your answer with the correct one above

Question

How many milliliters of a $5.0 M CuSO_{4}$ solution are needed to prepare of $0.500 M CuSO_{4}$ ?

Answer

of $5.0 M CuSO_{4}$ solution are needed to prepare of $0.500 M CuSO_{4}$ .

We can use the formula $M_{1}V_{1}= M_{2}V_{2}$

Therefore, $(5.00 M_{1})(V_{1})=(0.350 M_{2})(0.500 L)$

$V_{1}= 0.035 L = 35 mL$

Compare your answer with the correct one above

Question

What are the concentrations of aluminum and sulfate in a 3.0 M solution of aluminum sulfate?

Answer

The relative concentrations of aluminum sulfate $(Al_{2}(SO_{4})_{3})$ are

$6.0 M Al^{3+}$

$9.0 M SO_{4}^{2-}$

$\frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{2 mol Al^{3+}}{1 mol Al_{2}(SO_{4})_{3}}=6.0 M Al^{3+}$

$\frac{3.0 mol Al_{2}(SO_{4})_{3}}{1.0 L}\cdot\frac{3 mol SO_{4}^{2-}}{1 mol Al_{2}(SO_{4})_{3}}=9.0 M Al^{3+}$

Compare your answer with the correct one above

Question

What is the molarity of a solution containing of solution containing of ?

Answer

Molarity moles of solute per liter of solution.

We are given that there are of and of solution.

First lets convert the of solution to liters. It is easiest to use scientific notion in all calculations for easy simplification:

=

$\frac{7.510^{2}ml1L}{(10^3ml)}=7.510^{-1}L$

Now we must convert of to moles to get moles of solute per liter of solution.

This is done by dividing by its molecular weight which is . Once again we can use scientific notion to simplify calculations:

$3.510g\ MgCO_31mol\ MgCO_3\frac{1}{(8.41310g\ MgCO_3)}$

Now we can divide by $7.510^{-1}L$ solution to get the molarity.

$4.1610^{-1}mol\ MgCO_3\frac{1}{7.5*10^{-1}L}$

Compare your answer with the correct one above

Question

What is the molality of a solution made by adding of to of water?

Answer

Molality () is defined as moles of solute per kg of solvent.

is the solute (it is what is being dissolved) and water is the solvent (what is doing the dissolving).

Let’s start converting of to moles by dividing by its molecular weight using scientific notion through the entire molality calculations as necessary to simplify calculations.

$9.5g\ NaCl1mol\ NaCl\frac{1}{5.8410g\ NaCl}=0.163mol\ NaCl$

Now we must convert the of water to then divide by it to get the molality:

$310^{2}g\ H_2O1kg\frac{1}{10^3g}=310^{-1}kg\ H_2O$

Molality of solution:

$=1.6310^{-1}mol\ NaCl\frac{1}{310^{-1}kg\ H_2O}$

=

Compare your answer with the correct one above

Question

How many of water are needed to dilute to ?

Answer

In order to solve for the volume of water needed we must use this equation:

solve for since we are looking for the final volume of water needed to dilute the existing solution:

Compare your answer with the correct one above

Question

Which of the following choices is characteristic of molality?

Answer

Molarity, molality, and normality are all units of concentration in chemistry. Molarity () is defined as the number of moles of solute per liter of solution. Molality () is defined as the number of moles of solute per kilogram of solvent. Normality () is defined as the number of equivalents per liter of solution. Molality, as compared to molarity, is also more convenient to use in experiments with significant temperature changes. This is because the volume of a solution increases with temperature, and heating causes molarity to decrease; however, since molality is based on masses rather than volumes, molality remains unchanged.

Compare your answer with the correct one above

Question

A solution is prepared by dissolving $10.2\text{g of glucose,}\text{ C}_6\text{H}_{12}\text{O}_6\text{,}$ in $405\text{g}$ of water. The final volume of the solution is $414\text{mL}$ . Find the concentration of the solution in units of molality.

Answer

Recall how to find the molality of a solution:

$\text{Molality}=\frac{\text{moles of solute}}{\text{kg of solvent}}$

First, start by finding the moles of glucose that we have. The molar mass of glucose is $180.16\text{g/mol}$ .

$10.2\text{g of glucose}\cdot\frac{\text{1 mole glucose}}{180.16\text{g of glucose}}=0.0566\text{ moles of glucose}$

Next, convert the grams of water into kilograms.

$405\text{g}\cdot\frac{1\text{kg}}{1000\text{g}}=0.405\text{kg}$

Now, plug in the moles of glucose and kilograms of water into the equation for molality.

$\text{Molality}=\frac{0.0566}{0.405}=0.140m$

Compare your answer with the correct one above

Question

A solution of hydrogen peroxide is by mass. What is the molarity of the solution? Assume that the solution has a density of $1.01\text{g/mL}$ .

Answer

Start by assuming that we have $1.00\text{L}$ of this solution. Recall that hydrogen peroxide has a molecular formula of $\text{H}_2\text{O}_2$ .

Use the given density to find the mass of the solution.

$1.00\text{L}\cdot\frac{1000\text{mL}}{1\text{L}}\cdot\frac{1.01\text{g}}{\text{mL}}=1010\text{g}$

Next, find the mass of the hydrogen peroxide present in the solution.

$1010\text{g}(0.152)=153.52\text{g}$

Convert the mass of hydrogen peroxide into moles of hydrogen peroxide.

$153.52\text{g of H}_2\text{O}_2\cdot\frac{1\text{ mole H}_2\text{O}_2}{31.04\text{g H}_2\text{O}_2}=4.95\text{ moles of H}_2\text{O}_2$

Recall how to find the molarity of a solution:

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

Since we have $1.00\text{L}$ of solution, the molarity is .

Compare your answer with the correct one above

Tap the card to reveal the answer