Do I have to convert milliliters to liters in M_1V_1 = M_2V_2?

No. Volume appears on both sides of the equation, so the unit cancels as long as you use the same unit for V_1 and V_2. Milliliters are usually more convenient. You do need liters whenever you use M = n/V to find moles.

What is the difference between molarity and molality, and when do I use each?

Molarity is moles of solute per liter of solution; molality is moles of solute per kilogram of solvent. Use molarity for dilutions and for stoichiometry with volumes. Use molality for freezing-point depression and boiling-point elevation, because mass does not change with temperature while volume does.

Why does salt melt ice better than sugar?

Each formula unit of NaCl separates into two ions, while each sugar molecule stays as one particle. Freezing-point depression counts particles, so equal molalities of salt give roughly twice the depression of sugar. Calcium chloride is used on very cold roads because it releases three particles per formula unit.

Why is my calculated freezing point positive?

The equation T_f = i K_f m gives the size of the change, not the final temperature. Subtract it from the normal freezing point of the solvent: for water, 0.00\ ^ - T_f. A solution can never freeze above the pure solvent's freezing point.

CHEM-8.3

Dilutions & Colligative Properties

Learn to solve dilution problems with M1V1 = M2V2 and explain freezing-point depression and boiling-point elevation by counting dissolved solute particles.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Dilutions & Colligative Properties, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Dilutions & Colligative Properties free

What this lesson covers

Stockrooms almost never store the exact concentration you need. Instead they store concentrated stock solutions, and chemists dilute them on demand. Diluting sounds like it should be complicated, but it rests on one simple fact: adding water adds no solute. The moles of dissolved particles you started with are the moles you end with, so the concentration falls by exactly the factor the volume rises.

That same idea — counting dissolved particles — explains a second set of behaviors. Salt on an icy sidewalk, antifreeze in a radiator, and salt in pasta water all change the temperature at which a liquid freezes or boils. These are colligative properties, and what matters is how many particles dissolve, not what they are. In this lesson you will use

M_1V_1 = M_2V_2

fluently, and you will predict which solution freezes lowest and boils highest by counting ions.

Why $M_1V_1 = M_2V_2$ Works

Molarity is moles of solute per liter of solution,

M = \frac{n}{V}

. Rearranged,

n = MV

. When you dilute a solution you pour in more solvent, which changes

V

but not

n

— no solute enters or leaves the beaker. Setting the moles before equal to the moles after gives

M_1V_1 = M_2V_2

where subscript 1 is the concentrated stock and subscript 2 is the diluted solution.

Because the same volume unit appears on both sides, it cancels. You may use milliliters throughout as long as you use milliliters on both sides; there is no need to convert to liters. This is one of the few places in solution chemistry where mL is safe, and it saves time.

Two checks keep you honest. First, the diluted solution must always be less concentrated than the stock, so

M_2 < M_1

and

V_2 > V_1

. If your answer says otherwise, you swapped a pair of values. Second, the dilution factor works both ways: diluting from 6.0 M to 1.5 M is a factor of 4, so the final volume must be 4 times the starting volume.

A common mistake is reading

V_2

as the volume of water added. It is not.

V_2

is the total final volume of solution. The water added equals

V_2 - V_1

. If a problem asks how much water to add, subtract at the end.

The equation also works for dilutions expressed in other concentration units — parts per million, percent by mass — provided both sides use the same unit.

Performing a Dilution Correctly

Solving the math is half the task; the lab procedure matters too. Suppose you need 500.0 mL of 0.200 M sodium hydroxide from a 2.00 M stock. Then

V_1 = \frac{M_2V_2}{M_1} = \frac{(0.200)(500.0)}{2.00} = 50.0

mL of stock.

Step	Action	Why it matters
1	Calculate $V_1$ with $M_1V_1 = M_2V_2$	Tells you how much stock contains the needed moles
2	Measure $V_1$ with a pipet or graduated cylinder	Accuracy here sets the accuracy of the whole solution
3	Add the stock to a volumetric flask holding some water	Mixing order matters, especially for acids
4	Add solvent until the meniscus sits on the calibration line	The line marks total volume $V_2$ , not water added
5	Stopper and invert several times	Concentration must be uniform throughout

For concentrated acids, always add acid to water, never water to acid. Dilution of strong acids releases substantial heat, and adding water to concentrated acid can boil and spatter the mixture.

Students often go wrong by filling the flask to the line with water first and then adding the stock. That overshoots the final volume and makes the solution slightly too dilute in an amount you cannot measure or fix. The calibration mark on volumetric glassware refers to the finished solution, solute included.

Colligative Properties: Particles, Not Identity

A colligative property depends only on the concentration of dissolved solute particles, not on what those particles are. The two you will calculate are freezing-point depression and boiling-point elevation. Vapor-pressure lowering and osmotic pressure belong to the same family.

The mechanism for freezing-point depression is interference with crystal formation. To freeze, water molecules must lock into an ordered lattice. Dissolved particles get in the way and also make the liquid state more disordered and therefore more stable, so the liquid must be cooled below the normal freezing point before the solid can form. Boiling-point elevation comes from vapor pressure: solute particles at the surface reduce the number of solvent molecules that can escape, lowering the vapor pressure. A higher temperature is then required before vapor pressure equals atmospheric pressure.

Because only the count of particles matters, ionic compounds punch above their weight. One formula unit of

\text{NaCl}

dissolves into

\text{Na}^+

and

\text{Cl}^-

: two particles. One

\text{CaCl}_2

gives three. Glucose, a molecular compound, gives one. The van 't Hoff factor

i

is that particle count:

Solute	Dissolves into	$i$ (ideal)
Glucose, sucrose, ethanol	one molecule	1
$\text{NaCl}$ , $\text{KBr}$	2 ions	2
$\text{CaCl}_2$ , $\text{Na}_2\text{SO}_4$	3 ions	3
$\text{AlCl}_3$	4 ions	4

So 0.10 m

\text{CaCl}_2

depresses the freezing point about three times as much as 0.10 m glucose. Real values of

i

run slightly below the ideal because oppositely charged ions briefly pair up in solution, but the ideal number is what you use in this course.

Calculating $\Delta T_f$ and $\Delta T_b$

The working equations are

\Delta T_f = i \, K_f \, m \qquad \Delta T_b = i \, K_b \, m

Here

m

is molality, moles of solute per kilogram of solvent, not molarity. Molality is used because it does not change with temperature — mass does not expand when heated, but volume does, and these problems involve heating and cooling. In dilute aqueous solutions, molality and molarity are numerically close because one liter of dilute solution has a mass near one kilogram, so it is fine to approximate one from the other only when a problem tells you to.

For water,

K_f = 1.86\ ^\circ\text{C}/m

and

K_b = 0.512\ ^\circ\text{C}/m

. These constants belong to the solvent, not the solute; benzene and camphor have entirely different values.

The sign convention trips people up.

\Delta T_f

and

\Delta T_b

calculated from these formulas come out positive; you then apply them in the correct direction. Freezing point goes down: new freezing point

= 0.00\ ^\circ\text{C} - \Delta T_f

. Boiling point goes up: new boiling point

= 100.00\ ^\circ\text{C} + \Delta T_b

. Writing a freezing point of positive 3.7 degrees Celsius for salt water is the single most common error on this material.

Two more places students slip. First, forgetting

i

entirely for an ionic solute, which makes the answer too small by a factor of 2 or 3. Second, dividing by the mass of the whole solution instead of the mass of the solvent alone when finding molality. The denominator of molality is the solvent mass by itself, so the mass of dissolved solute never belongs in it.

Key terms

Dilution.: The process of adding solvent to a solution to lower its concentration; the number of moles of solute stays constant.
Stock solution.: A concentrated solution kept on hand and diluted to make the working concentrations needed for an experiment.
Molarity ( $M$ ).: Moles of solute per liter of solution. Used in $M_1V_1 = M_2V_2$ .
Molality ( $m$ ).: Moles of solute per kilogram of solvent. Used in colligative-property equations because it is independent of temperature.
Colligative property.: A solution property that depends only on the number of dissolved solute particles, not their chemical identity.
Freezing-point depression.: The lowering of a solvent's freezing point by dissolved solute, calculated as $\Delta T_f = i K_f m$ .
Boiling-point elevation.: The raising of a solvent's boiling point by dissolved solute, calculated as $\Delta T_b = i K_b m$ .
van 't Hoff factor ( $i$ ).: The number of particles one formula unit of solute produces in solution: 1 for molecular solutes, 2 for NaCl, 3 for CaCl₂.

Worked example

You need 250.0 mL of 0.150 M NaCl for a lab, and the stockroom has 2.00 M NaCl. (a) What volume of stock is required, and how much water do you add? (b) Assuming the dilute solution has a density close to water so its molality is essentially 0.150 m, find its freezing point. For water,

K_f = 1.86\ ^\circ\text{C}/m

Part (a). Identify the four quantities:

M_1 = 2.00

M_2 = 0.150

V_2 = 250.0

mL, and

V_1

is unknown. Because the volume unit cancels, work in milliliters.

V_1 = \frac{M_2 V_2}{M_1} = \frac{(0.150\ \text{M})(250.0\ \text{mL})}{2.00\ \text{M}} = 18.75 \approx 18.8\ \text{mL}

Check the direction: the stock is about 13 times more concentrated, and 250.0 mL is about 13 times 18.8 mL. Consistent.

Water added

= V_2 - V_1 = 250.0 - 18.8 = 231.2

mL. In practice you measure 18.8 mL of stock into a 250.0 mL volumetric flask and fill to the mark, which accomplishes the same thing without measuring the water separately.

Part (b). NaCl is ionic and dissociates into

\text{Na}^+

and

\text{Cl}^-

, so

i = 2

\Delta T_f = i K_f m = (2)(1.86\ ^\circ\text{C}/m)(0.150\ m) = 0.558\ ^\circ\text{C}

Apply it in the correct direction — freezing points go down:

T_f = 0.000\ ^\circ\text{C} - 0.558\ ^\circ\text{C} = -0.558\ ^\circ\text{C}

The solution freezes at about negative 0.56 degrees Celsius. Sanity check: a dilute solution should freeze only slightly below zero, and forgetting the factor

i

would have given half this depression.

Practice questions

A student dilutes 50.0 mL of 6.0 M HCl to a final volume of 300.0 mL. What is the concentration of the diluted acid?

0.50 M
1.0 M
1.2 M
36 M

Answer: 1.0 M

Use

M_2 = \frac{M_1V_1}{V_2} = \frac{(6.0)(50.0)}{300.0} = 1.0

M. The volume increased by a factor of 6, so the concentration must drop by a factor of 6. Choosing 36 M means the numbers were multiplied instead of divided, and it fails the basic check that a diluted solution is always less concentrated than the stock. Choosing 1.2 M comes from dividing by 250 mL, the water added, rather than by the total final volume.

Rank these three aqueous solutions from highest freezing point to lowest: 0.20 m glucose, 0.20 m NaCl, 0.20 m CaCl₂. Explain your reasoning using the number of dissolved particles.

Answer: Highest to lowest freezing point: 0.20 m glucose, then 0.20 m NaCl, then 0.20 m CaCl₂.

Freezing-point depression depends on total particle concentration, so multiply molality by

i

. Glucose is molecular,

i = 1

, giving 0.20 m particles and

\Delta T_f = (1)(1.86)(0.20) = 0.37\ ^\circ\text{C}

, so it freezes at about negative 0.37 degrees Celsius. NaCl gives 2 ions, so 0.40 m particles and

\Delta T_f = 0.74\ ^\circ\text{C}

, freezing near negative 0.74 degrees Celsius. CaCl₂ gives 3 ions, so 0.60 m particles and

\Delta T_f = 1.12\ ^\circ\text{C}

, freezing near negative 1.12 degrees Celsius. The greatest depression means the lowest freezing point, so CaCl₂ is last in the ranking. Notice all three have the same molality — only the particle count differs.

What volume of 12.0 M stock ammonia solution is needed to prepare 2.00 L of 0.500 M ammonia, and briefly describe how you would prepare it?

Answer: 83.3 mL of stock, added to a 2.00 L volumetric flask and diluted to the mark with water.

Convert nothing: keep volumes consistent. Using liters,

V_1 = \frac{M_2V_2}{M_1} = \frac{(0.500)(2.00)}{12.0} = 0.0833

L, or 83.3 mL. To prepare it, add some water to a 2.00 L volumetric flask, add 83.3 mL of the stock, then add water until the meniscus rests on the 2.00 L calibration line, and invert repeatedly to mix. The final line marks total solution volume, so you do not measure out 1.917 L of water separately.

FAQ

Do I have to convert milliliters to liters in $M_1V_1 = M_2V_2$ ?: No. Volume appears on both sides of the equation, so the unit cancels as long as you use the same unit for $V_1$ and $V_2$ . Milliliters are usually more convenient. You do need liters whenever you use $M = \frac{n}{V}$ to find moles.
What is the difference between molarity and molality, and when do I use each?: Molarity is moles of solute per liter of solution; molality is moles of solute per kilogram of solvent. Use molarity for dilutions and for stoichiometry with volumes. Use molality for freezing-point depression and boiling-point elevation, because mass does not change with temperature while volume does.
Why does salt melt ice better than sugar?: Each formula unit of NaCl separates into two ions, while each sugar molecule stays as one particle. Freezing-point depression counts particles, so equal molalities of salt give roughly twice the depression of sugar. Calcium chloride is used on very cold roads because it releases three particles per formula unit.
Why is my calculated freezing point positive?: The equation $\Delta T_f = i K_f m$ gives the size of the change, not the final temperature. Subtract it from the normal freezing point of the solvent: for water, $0.00\ ^\circ\text{C} - \Delta T_f$ . A solution can never freeze above the pure solvent's freezing point.

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