#### Practice

Identify the solvent and solute(s) in a sample of an antiseptic that contains 65% propanol, 20% ethanol, and 15% water.

Solvent – propanol

Solutes – ethanol and water

A solute is the material present in the smaller amount in the solution.

The solvent is the material present in the larger amount in the solution.

What is the molarity of the solution prepared by dissolving 33.5 g sodium oxalate (Na_{2}C_{2}O_{4}) in 250.0 ml water?

1.0 mol/L

The molarity is calculated by the following formula:

M = n (mol)/V (L)

First, determines the moles of sodium oxalate:

n (Na_{2}C_{2}O_{4}) = 33.5 g / 134 g/mol = 0.25 mol

Next, convert the mL to L in order to match the formula.

V = 250 mL x 1 L/1000 mL = 0.25 L

And now, plug in the numbers to determine the concentration of sodium oxalate:

**M = n (mol)/V (L) = 0.25 mol/0.25 L = 1.0 mol/L**

How many grams of silver nitrate (AgNO_{3}) can be obtained from an 800.0 g sample of a 65% solution?

520 g AgNO_{3}

65% solution means there is 65 g AgNO_{3 }in 100 g solution. Therefore, in 800.0 g, there is 800.0 x 065 = 520 g AgNO_{3.}

Commercial concentrated hydrochloric acid is 36.0% by mass with a density of 1.2 g/mL. Calculate the molarity of this solution.

11.9 mol/L

The molarity is calculated by the following formula:

M = n (mol)/V (L)

So, first, we need to determine the moles of hydrochloric acid. If the quantity of the solution is not given, you can assume any number and for convenience, 100 g solution is a good number. In the 100 g solution, there will be 36.0 g HCl. Therefore, the moles of HCl will be:

n (HCl) = m (g) / M (g/mol) = 36.0 g / 36.5 g/mol = 0.987 mol.

Next, we need to determine the volume of the solution which can be calculated using the mass and the density of the solution.

V = m (g) / d (g/mL) = 100.0 g / 1.2 g/mL = 83 mL

We also need to convert this to L:

V (L) = 83 mL x 1 L/1000 mL = 0.083 L

And last, the molarity will be:

M = n (mol)/V (L) = 0987 mol / 0.083 L = 11.9 mol/L

A student has 2.0 L of 7.0 M hydrochloric acid, HCl. She wants to prepare a 1.5 M solution. How many liters of solution can she make?

9.3 L

To find the volume of the solution, we need to use the equation for dilution:

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

Next, assign the volume and concentration for the initial and final solutions. It is intuitive to assign M_{1 }and V_{1 }for the initial (stock) solution, and M_{2, }V_{2 }for the final solution.

We are looking for V_{2} and therefore, we need to **rearrange the equation**:

V_{2 }= M_{1}V_{1} / M_{2 }= 7.0 M x 2.0 L / 1.5 M = **9.3 L**

What is the concentration of the solution made by mixing 100. mL of 2.0 M and 250. mL 3.5 M solutions of potassium iodide?

** 3.1 mol/L or 3.1 M**

This is a problem that can be solved by the dilution formula:

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

The only difference from these types of common questions is that here we have two stock solutions. So, M_{1, }V_{1, }and M_{2, }V_{2 }are all for initial concentrations and we need to assign M_{3, }V_{3 }for the final solution. Therefore,

M_{1}V_{1} + M_{2}V_{2 }= M_{3}V_{3}

One last thing before plugging in the numbers is to convert to mL to L.

V_{1 }= 100. mL x 1 L/1000 mL = 0.100 L

V_{2 }= 250. mL x 1 L/1000 mL = 0.250 L

V_{3 }= 0.100 L + 0.250 L = 0.350 L

Next, rearrange the dilution equation to calculate V_{3}:

M_{3 }= (M_{1}V_{1} + M_{2}V_{2}) / V_{3 }= (2.0 M x 0.100 L + 3.5 M x 0.250 L) / 0.350 L =** 3.1 mol/L or 3.1 M**

How many grams of copper(II) sulfate pentahydrate (CuSO_{4}·5H_{2}O) is needed to prepare 500. mL solution of 0.480 M CuSO_{4}(aq)?

**59.9 g CuSO _{4}·5H_{2}O**

The molarity is given by the following formula:

M = n (mol)/V (L)

We are given the molarity (0.480 M) and V (500. mL) and we need to find the mass. The link between the formula of molarity and mass is the moles. Therefore, we first need to calculate the moles of CuSO_{4}·5H_{2}O. For this, rearrange the equation:

n (mol) = M x V (L) = 0.480 M x 0.500 L = 0.240 mol CuSO_{4}

Now, these are the moles of CuSO_{4, }but we are asked to find the needed mass of CuSO_{4}·5H_{2}O. Although the molar masses of these salts are different, their moles are the same because each mol of CuSO_{4}·5H_{2}O contains 1 mol CuSO_{4. }Therefore_{, }the moles of CuSO_{4}·5H_{2}O needed to prepare the solution are the same as the moles of CuSO_{4}. In other words, if we need to supply 5 moles of CuSO_{4}, then 5 moles of CuSO_{4}·5H_{2}O are going to be needed.

So, in this case, n(CuSO_{4}·5H_{2}O) = n(CuSO4) = 0.240 mol

The mass is calculated from the equation

n (mol) = m (g) /M (g/mol)

therefore,

m (g) = n (mol) x M (g/mol) = 0.240 mol x 249.7 =** 59.9 g CuSO _{4}·5H_{2}O**

An ICU nurse is preparing for an intravenous administration of glucose (C_{6}H_{12}O_{6}). How many mL of this solution will be needed to provide 6.50 mmol of glucose if the solution is labeled as 0.350 M?

**19 mL**

Like the problems before, we need to use the formula for molarity. One difference here is that the amounts are in millimoles, so we need to do a conversion first.

Moles of glucose = 6.50 mmol x 1 mol/1000 mmol = 0.0065 mol

From here, we can find the volume of solution in liters and convert to mL.

V = n/M = 0.0065 mol/0.350 M = 0.019 L

V (mL) = 0.019 L x 1000 mL/1L = **19 mL**

How many milliliters of 0.850 M BaCl_{2} solution are needed to obtain 9.37 g of the salt?

**52.9 mL**

First, covert the mass of BaCl_{2} to moles.

n = m/M = 9.37 g / 208.2 g/mol = 0.0450 mol

Next, we can find the volume using the equation for molarity:

V = n/M = 0.0450 mol / 0.850 M = 0.0529 L

And now, convert the liters to milliliters:

V (mL) = 0.0529 L x 1000 mL/1L = **52.9 mL**

Calculate the concentration of each ion and their total concentration in a solution prepared by dissolving 0.2500 mol of Ca(NO_{3})_{2} in 150.0 mL of water.

**M (Ca ^{2+}) ^{ }= 1.667 mol/L**

**M (NO _{3}^{–}) ** =

**3.334**

**Total concentration of all ions** = **5.0 M**

First, let’s write down the dissociate equation of Ca(NO_{3})_{2}:

Ca(NO_{3})_{2 }→ Ca^{2+} + 2NO_{3}^{–}

Because each mole of Ca(NO_{3})_{2 }produces one mol of Ca^{2+}, their concentration is also going to be the same. To find the concentration of Ca(NO_{3})_{2, }convert the volume from mL to L and use in the formula for molarity:

V (L)_{ }= 150.0 mL x 1 L/1000 mL = 0.1500 L

M (Ca(NO_{3})_{2}) = n/V = 0.2500 mol / 0.1500 L = 1.667 mol/L

Therefore, the **concentration of Ca ^{2+ }ions is also 1.667 mol/L**.

On the other hand, each mole of Ca(NO_{3})_{2 }produces 2 moles of NO_{3}^{– }ions. So, the **concentration of NO _{3}^{– }ions** = 2 x 1.667 mol/L =

**3.334 mol/L.**

The **total concentration of all ions** = M (Ca^{2+}) + M(NO_{3}^{–}) = 1.667 M + 3.334 M = **5.0 M**

Calculate the molality of a 0.500 M solution (d = 1.09 g/mL) of baking soda (NaHCO_{3}).

**0.476 mol/kg**

The molality is calculated by the formula **m = mol solute / kg solvent.**

Because we are not given any numbers for the amount of solute, solvent, or solution, we can assume any amount. Let’s assume we have a 1.00 L of solution. In this case, the **moles of NaHCO**_{3 }will be the same as its molarity – **0.500 mol**.

So, the only missing number now is the mass of the solvent in kilograms. To find this, we can calculate the mass of the solute and subtract that from the mass of the solution.

The **mass of NaHCO _{3 }**= 0.500 mol x 84.0 g/mol = 42.0 g =

**0.0420 kg**

The mass of the solution can be calculated using the volume and density of the solution. We need to convert the volume to mL since the units for density are given as g/mL.

V (mL) = 1.00 L x 1000 mL/1L = 1000. mL

m = d x V = 1.09 g/mL x 1000. mL = 1090. g solution

This, in turn, needs to be converted to kg.

m (kg) = 1090 g x 1 kg/1000g = 1.09 kg

Therefore, the **mass of solvent** = 1.09 kg – 0.0420 kg** = 1.05 kg**

Last, the molality is:

m = 0.500 mol / 1.05 kg =** 0.476 mol/kg**

A solution contains 3.5 ×10^{–3} g of vitamin C in 1,625 mL of solution. What is the concentration of vitamin C in parts per million (ppm) if the density of the solution is 1.08 g/mL.

** 2.0 ppm**

The concentration in parts per million (ppm) is expressed by dividing the amount of solute over the amount of solution, in the same units, and multiplying the result by 10^{6}. Because the amount of vitamin C is given in grams, we need to determine the mass of the solution in grams as well.

The mass of the solution can be calculated using the volume and density of the solution.

m = d x V = 1.08 g/mL x 1,625 mL = **1,755 g solution**

ppm = (3.5 ×10^{–3} g of vitamin C/1,755 g solution) x 10^{6} =** 2.0 ppm**

Calculate the *mass percent, mole fraction, molality, *and *molarity* of hexane (C_{6}H_{14}, *d *=0.660 g/cm^{3}) when it is mixed with heptane (C_{7}H_{12}, *d=*0.684 g/cm3) in a 35.0 : 55.0 mL ratio. Assume that the volumes are additive.

**% mass** = **38.1 %**

**Mole fraction ****= 0.420**

**Molality ** = **0.0376 mol/kg**

**Molarity** = **2.98 mol/L**

Let’s start with the **mass percent**. It is the ratio of the mass of hexane over the mass of the solution multiplied by 100 %.

The mass of hexane is:

**mass (hexane)** = d x V = 0.660 g/mL^{ }x 35.0 mL =** 23.1 g**

Notice that 1 cm^{3 }= 1 mL, and therefore, 0.660 g/cm^{3 }= 0.660 g/mL.

Next, find the mass of heptane.

**mass (heptane)** = d x V = 0.684 g/mL^{ }x 55.0 mL = **37.6 g**

The **mass of the solutuion** is 23.1 g + 37.6 g =** 60.7 g**

**% (hexane)** = (23.1 g / 60.7 g) x 100% = **38.1 %**

**The mole fraction** is the ratio of the moles of hexane over the moles of the solution.

Moles of hexnae = m/M = 23.1 g/86.2 g/mol = 0.268 mol

Moles of heptane = m/M = 37.6 g/100.2 g/mol = 0.375 mol

Total moles = 0.268 mol + 0.375 mol = 0.643 mol

**Mole fraction of hexane = 0.268 mol/0.643 mol = 0.420**

**The molality of hexane** is:

m (hexne) = moles hexane/kg heptane

The mass of heptane in kilograms is 37.6 g x 1 kg / 1000 g = **0.0376 mol/kg**

Therefore, the molality of hexane is:

0.268 mol / 0.0376 kg **= 7.13 mol/kg**

**The molarity** of hexane is M (hexane) = n/V

The volume of the solution is 55.0 + 35.0 = 90.0 mL = 0.0900 L

**M (hexane)** = n/V = 0.268 mol / 0.0900 L = **2.98 mol/L**

Ethylene glycol (C_{2}H_{6}O_{2}) is used in automobile antifreeze as a 40.0 mass % aqueous solution. Calculate the molarity, molality, and mole fraction of the ethylene glycol if the density of the solution is 1.05 g/cm^{3}.

**The molarity** = **6.76 mol/L**

**The molality ****= 10.7 mol/kg**

**Mole fraction = 0.162**

If the quantity of the solution is not given, you can assume any number. Let’s set to a 1 L solution this time. That is 1000 mL or 1000 cm^{3}.

**The molarity** of ethylene glycol is M = n/V. We have the volume, so next is to find the moles by determining the mass first.

The mass of the solution will be:

m = d x V = 1.05 g/cm^{3 }x 1000 cm^{3} = 1050 g

In this solution, we have 40.0 % ethylene glycol which is calculated as follows:

m (C_{2}H_{6}O_{2}) = m (solution) x 0.400 = 1050 g x 0.4 = 420. g

The moles of ethylene glycol is found by the ratio of the mass over the molar mass:

n = m/M = 420. g / 62.1 g/mol = 6.76 mol

**The molarity** of ethylene glycol is M = n/V = 6.76 mol/1 L = **6.76 mol/L**

**The molality of ethylene glycol **is:

m (ethylene glycol) = moles C_{2}H_{6}O_{2}/kg H_{2}O

The mass of water (the solvent) is equal to the mass (solution) – mass (C_{2}H_{6}O_{2}) = 1050 g – 420. g = 630. g = 0.630 kg

Therefore, the molality of ethylene glycol is:

6.76 mol / 0.630 kg **= 10.7 mol/kg**

**The mole fraction** is the ratio of the moles of ethylene glycol over the moles of the solution (total moles of C_{2}H_{6}O_{2 }and H_{2}O).

Moles of water = m/M = 630. g/18.0 g/mol = 35.0 mol

Total moles = 6.76 mol + 35.0 = 41.8 mol

**Mole fraction of ethylene glycol = 6.76 mol/41.8 mol = 0.162**

A solution containing 186.5 g of KCl in 813.5 g of water has a density of 1.106 g/mL. What is the molarity of the solution?

**The molarity** = **2.76 mol/L**

**The molarity** is calculated by the formula M = n/V. So first, let’s find the moles of KCl by the ratio of the mass over the molar mass:

n = m/M = 186.5 g / 74.6 g/mol = 2.50 mol

Next, calculate the volume of the solution using its mass and density.

The mass of the solution is 186.5 g + 813.5 g = 1000.0 g

The volume of the solution will be:

V = m / d = 1000.0 g/1.106 g/mL = 904.2 mL

Convert the volume to liters to match the formula for molarity.

V (L) = 904.2 mL x 1L/1000 mL = 0.9042 L

**The molarity** of KCl is M = n/V = 2.50 mol/0.9042 L = **2.76 mol/L**

A lab specialist has 20% and 60% stock solutions of sodium hydroxide. Suppose she is asked to prepare 100 g of 35% solution. How many grams of each solution will she need to mix?

They need to mix 62.5 g of the 20% solution with 37.5 g of the 60% solution.

This is going to be two unknowns with two equations.

Let’s assign x g of 20% and y g of the 60% solutions that are needed to prepare the 35% solution. The first equation is set according to the mass of the final solution which needs to be 100 g. Now, because x and y are going to make this solution, we can say that their sum is equal to 100. So, the first equation is:

**x + y = 100**

For the second equation, we need to determine the masses of sodium hydroxide in solutions x and y. In x g of solution, there is 0.2x g of sodium hydroxide because 0.2x is the 20% of x.

Consequently, in y g solution, there is 0.6y g of NaOH. We also know that the total mass of NaOH must be 35 g in the final solution, and therefore, we can set up the second equation:

**0.2x + 0.6y = 35**

Using the first equation, we get that x = 100 – y. And we can plug this value in the second equation:

**0.2(100 – y) + 0.6y = 35**

20 -0.2y + 0.6y = 35

**y = 37.5**

Therefore, **x = 100 – 37.5 = 62.5 g**

So, we need to mix 62.5 g of the 20% solution with 37.5 g of the 60% solution.

**Alternatively**, for the second equation, we could have said that:

**(0.2x + 0.6y) / x + y = 0.35**

This is the expression for the percent concentration and solving it together with the x + y = 100 equation will give the same numbers.

What is the molality of the solution prepared by mixing 625.2 grams of menthol (C_{10}H_{20}O) and 3,500. grams of ethanol (C_{2}H_{5}OH).

**1.143 mol/kg**

The molality is calculated by the formula **m = mol solute / kg solvent.**

Because menthol is present in less quantity, it is the solute, and ethanol is the solvent. So, we need the moles of menthol and the mass of ethanol in kilograms.

n (menthol ) = m/M = 625.2 g / 156.3 g/mol = 4.000 mol

The **mass of ethanol **= 3,500. g x 1 kg/1000 g = **3.500 kg**

Last, the molality is:

m = 4.000 mol / 3.500 kg =** 1.143 mol/kg**

Calculate the molality of 31.0% MgCl_{2} solution with a density of 1.27 g/mL.

**4.7 mol/kg**

If the quantity of the solution is not given, you can assume any number. Let’s assume we have a 1 kg (1000 g) solution.

This means, there are 0.310 x 1000 g = 310. g MgCl_{2}

n (MgCl_{2}) = m/M = 310. g / 95.2 g/mol = 3.25 mol

The solvent here is presumably water since the solute is an inorganic salt. To find the mass of the solvent, we subtract the mass of the solute from the mass of the solution.

m (solvent) = 1000 g – 310. = 690 g = 0.69 kg

The molality is:

**m = 3.25 mol / 0.69 kg = 4.7 mol/kg**

How many grams of barium chloride (BaCl_{2}) are in a 0.025 molal solution if 3.50 kg of solvent was used to prepare it?

**18.2 g**

The molality is calculated by the ratio of the moles of BaCl_{2 }and the mass of solvent in kilograms:

m = moles (BaCl_{2}) / kg solvent

We have the mass of the solvent and the molality. To find the mass of BaCl_{2 }needed to prepare the solution, we can determine the moles from the formula for molality and convert them to the mass.

moles (BaCl_{2}) = m (molality) x kg solvent = 0.025 mol/kg x 3.50 kg = 0.0875 moles

mass (BaCl_{2}) = 0.0875 mol x 208.2 g/mol = 18.2175 = **18.2 g**

Determine the percentage concentration of 5.00 M HCl solution if its density is 1.104 g/mL.

**3.31 %**

Percentage concentration is the ratio of the mass of HCl over the mass of the solution multiplied by 100 %. It tells us how many grams of solute there are in a 100 g solution.

Let’s assume we have a 1.00 L solution which sets the moles of HCl to 1. So, there is 1 .00 mol of HCl in this solution. The mass of HCl is:

m (HCl) = 1.00 x 36.5 g/mol = 36.5 g

The mass of the solution is

**mass (solution)** = d x V = 1.104 g/mL^{ }x 1000. mL =** 1104 g**

**% (HCl )** = (36.5 g / **1104 ** g) x 100% = **3.3061 % = 3.31 %**

Glycerol (CH_{3}CH(OH)CH_{2}OH) is widely used in FDA-approved wound and burn treatments. What volume of glycerol (d = 1.26 g/mL), must be added to one kilogram of water to produce a solution with 5.00 mass % glycerol?

**41.7 mL**

Let’s set the following abbreviations:

m(g) = mass of glycerol

m(w) = mass of water

m(s) = mass of solution

The percent mass of glycerol is the ratio of its mass over the mass of the solution:

% (glycerol) = m (g) / [m(g) + m(w)]

So, we can set up an equation with one unknown which is the mass of glycerol:

% (glycerol) = m (g) / [m(g) + m(w)] = 0.05 (5 %)

0.05 = m(g)/ (m(g) + 1000)

0.05 m(g) + 50 = m(g)

**m(g) = 52.6 g**

Next, calculate the volume of glycerol using its density:

V = m/d = 52.6 g/1.26 g/mL = **41.7 mL**

A bottle of rum contains 40% ethanol by volume. The density of ethanol (C_{2}H_{5}OH) is 0.789 g/cm^{3}. Calculate the mass percent and molality of ethanol in the rum.

**mass % (C _{2}H_{5}OH) = 34.5 %**

**The molality of ethanol ****= 1.14 mol/kg**

The first thing here is to realize that water makes the other 60% of the solution. So, assuming a 100.0 mL of solution, we know that there are 40 mL of ethanol and 60 mL of water. To find the mass % of ethanol we need the masses of ethanol and water.

m(C_{2}H_{5}OH) = d x V = 0.789 g/cm^{3 }x 40.0 cm^{3 }= 31.6 g

m (H_{2}O) = d x V = 1.00 g/cm^{3 }x 60.0 cm^{3 }= 60.0 g

**mass % (C _{2}H_{5}OH) = 31.6 g/ (31.6 g + 60.0 g) = 34.5 %**

The formula of molality is:

m (C_{2}H_{5}OH) = moles C_{2}H_{5}OH/kg H_{2}O

The mass of water is kg is

60.0 g x 1 kg/1000g = 0.600 kg

The moles of ethanol = 31.6 g/46.1 g/mol = 0.685 mol

Therefore,** the molality of ethanol is:**

**0.685 mol / 0.600 kg = 1.14 mol/kg**

15.00 mL of propanol (CH_{3}CH_{2}CH_{2}OH) was dissolved in sufficient volume of water to prepare 100.0 mL of a solution with a density of 0.983 g/mL. Given the density of propanol is 0.803 g/mL, determine its concentration expressed as **(a) **volume percent; **(b) **mass percent; **(c) **mass/volume percent; **(d) **mole fraction; **(e) **mole percent; **(f) **molarity; **(g) **molality?

**(a) **volume percent** = 15.00%**

**(b) **mass percent** = 12.3% **

**(c) **mass/volume percent **= 12.05 %**

**(d) **mole fraction** = 0.0403**

**(e) **mole percent** = 40.3%**

**(f) **molarity** = 2.00 mol/L**

**(g) **molality** = 2.32 mol/kg**

**(a) **volume percent:

This is the ratio of the propanol volume over the volume of the solution:

**% volume (CH _{3}CH_{2}CH_{2}OH) = 15.00 mL/100.0 mL x 100% = 15.00%**

**(b) **mass percent:

This is the ratio of the propanol mass over the mass of the solution.

mass propanol = 15.00 mL x 0.803 g/mL = 12.045 g

mass solution = 100.0 mL x 0.983 g/mL = 98.3 g

**% mass propanol = 12.045 g / 98.3 g x 100% = 12.3% **

**(c) **mass/volume percent

This is the ratio of the propanol mass over the volume of the solution.

% mass/volume = **12.045 g /100.0 mL x 100% = 12.05 %**

**(d) **mole fraction

This is the ratio of the propanol moles over the total moles of propanol and water.

Convert the mass of propanol to moles:

n (propanol) = m/M = 12.045 g/60.1 g/mol = 0.200 mol

mass water = mass solution – mass propanol = 98.3 g – 12.045 g = 86.3 g

Convert the mass of water to moles:

n (water) = m/M = 86.3 g/18.0 g/mol = 4.76 mol

**χ (propanol) = 0.200 mol/0.200 mol + 4.76 mol = 0.200 mol/4.96 mol = 0.0403**

**(e) **mole percent;

This is the mole fraction multiplied by 100%.

**mole percent = 0.0403 x 100% = 40.3%**

**(f) **molarity;

The molarity is the ratio of the moles over the volume of solution in liters.

**M (propanol) = n/V = 0.200 mol/0.1000 L = 2.00 mol/L**

**(g) **molality

This is the ratio of the moles over the mass of solvent (water) in kilograms.

mass water = 86.3 g x 1kg/1000 g = 0.0863 kg

**molality (propanol) = 0.200 mol/0.0863 kg = 2.32 mol/kg**