Menu
## Dilution and Density of Solutions

**Dilution and Density of Solutions**

Dilution is process of adding solvent to solution. Since amount of solute stays constant, concentration of solution decreases. We find relation between concentration of solutions before and after dilution with following formula;

**M _{1}.V_{1}=M_{2}.V_{2}**

Where M_{1} is initial molarity and M_{2} is final molarity and V_{1} and V_{2} are initial and final volumes of solution.

To increase concentration of solutions, you should add solute or evaporate solvent from solution. Formula given above is also used in increasing concentration of solutions;

**M _{1}.V_{1}=M_{2}.V_{2}**

Concentration of solutions and volumes are inversely proportional to each other. If volume of solution increases then, molarity of solution decreases. Graph given below shows this relation;

**Example:** If we add 700 mL water at same temperature to 0,2 molar 300 mL solution, find final molar concentration of this solution.

**Solution: **

M_{1}=0,2 molar, V_{1}=300=0,3 mL

V_{2}=300+700 =1000mL=1 L

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

0,2.0,3=M_{2}.1

M_{2}=0,06 molar

**Example:** If we mix solutions given in the picture below, find concentration of final solution.

**Solution: **Sum of masses solution one and two gives us mass of final solution

m_{1}+m_{2}=m_{final}

80 + 40 =120 g

Sum of solute masses one and two gives us mass of final solute.

**(1) **m_{sugar1} + m_{sugar2}=m_{sugarf}

We find masses of solutes by;

m_{sugar1}=m_{1}.20/100=80.1/5=16 g

m_{sugar2}=m_{2}.30/100=40.30/100=12 g

m_{sugarf}=m_{final}.X/100=120.X/100 g

we use equation (1) and solve for X;

m_{sugar1} + m_{sugar2}=m_{sugarf}

16 + 12=120.X/100 g

28=12.X/10

X=23,3

**Density of Solutions**

We find density of solutions by following formula;

Unit of liquid solutions g/mL or g/cm^{3}. Putting solute into water we prepare solution. When we add solute to solution density of it increases, since increase in the mass of solution is larger than the increase in volume. In solid-liquid solutions, density increases with increasing in the concentration of solution.

**Example:** Density of H_{2}SO_{4} solution, having percent by mass 49 %, is 1,2 g/mL. Find molar concentration of this solution. (H_{2}SO_{4}=98)

**Solution: **

density of solution=1,2 g/mL

Percent by mass= 49 %

Molar mass of H_{2}SO_{4} is 98 g

We find molar concentration of solution with following formula;

M=(1,2.4)/98 . 1000

M=6 molar

**Example:** Solubility of X at 15 ^{o}C is 20g X/100. Which ones of the following statements are true for solution prepared using 30 g X and 120 g water at 15 ^{o}C?

**I.** Solution is saturated.

**II.** Mass of solution is 150 g.

**III. **Concentration percent by mass is 20 %

**Solution: **

**I.** at 15^{o}C

100 g water dissolves 20 g X

120 g water dissolves ? g X

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

?=24 g X is dissolved.

since 30 g X is added to 120 g water, solution is saturated and 30-24=6 g X stays undissolved. I is true.

**II.** Mass of solution is equal to sum of solute and solvent.

m=msolute+msolvent

m=120+24=144

Thus mass is not equal to 150 g, II is false.

**III.** Since 100 g water dissolves 20 g X, there are also 20 g X in 120 g solution. Thus, percent by mass;

X %=(m_{X}/msolution).100

X %=(20/120).100=16,7

III is false

The Original Author: Mrs. Şerife (Erden) SARICA

© Copyright www.ChemistryTutorials.org, Reproduction in electronic and written form is expressly forbidden without written permission of www.ChemistryTutorials.org. Privacy Policy