WORKSHEET ON SOLUTION CONCENTRATIONS

Worksheet 10.8

worksheet on solution concentrations

introduction

A solution is made up of two parts, the SOLVENT and the SOLUTE. The solution could be CONCENTRATED or DILUTE.

SOLVENT

The liquid which dissolves the solute. It is usually water.

SOLUTE

The substance which dissolves - often a solid, but not always.

CONCENTRATEDThere is a lot of solute and a little water.

DILUTE

There is a lot of water and only a little solute.

When we measure out 50cm3 of a solution, e.g., hydrochloric acid, we measure out some solvent (water) and some acid together. In any reactions of the acid, it is the HCl particles which are involved. The water is just a carrier for the acid, so when we measure out a volume of the solution, we want to know how much acid it contains. To do this, we need to define the CONCENTRATION of the solution. This tells us how much solute there is in a fixed volume of the solution. In chemistry, how much means MOLES and the fixed volume is taken as 1 cubic decimetre (1dm3).

[NB: 1dm3 is 1000cm3 and is the same volume as 1 litre.]

definition

You must learn this relationship and be able to use it and rearrange it to calculate moles of solute or volume of solution, given the other two quantities.

Remember that the volume must be in dm3, so convert from cm3 if necessary.

The units for concentration are moles per cubic decimetre which is written mol.dm-3 Lab bottles use the letter M (meaning molar) and this is the same thing as mol.dm-3

relationships

Use the triangle opposite to help you answer the questions which follow. Some worked examples are given to help you.

Examples:-

1) What is the concentration of a solution which contains 2 moles of copper (II) sulphate in 4dm3 of solution?

Ans:concentration = moles ( volume = 2 ( 4 = 0.5 mol.dm-32) How many moles of sodium hydroxide are there in 200cm3 of a solution of concentration 0.2 mol.dm-3 ? (Remember that 200cm3 = 200 ( 1000 = 0.2dm3)

Ans:moles = concentration ( volume = 0.2 ( 0.2 = 0.04 molesquestions

3) Calculate the concentration in mol.dm-3 of each of the following solutions:-

a) 10 moles of potassium hydroxide in 5dm3 of solution.

b) 1.5 moles of hydrochloric acid in 3dm3 of solution.

c) 0.5 moles of calcium chloride in 500cm3 of solution.

d) 0.2 moles of sulphuric acid in 100cm3 of solution.

e) 0.075 moles of magnesium sulphate in 25cm3 of solution.

4) Find the number of moles of solute in each of the following solutions:-

a) 2dm3 of 2 mol.dm-3 nitric acid.

b) 0.25dm3 of 0.4 mol.dm-3 ammonium chloride solution.

c) 200cm3 of 0.8 mol.dm-3 sodium carbonate solution.

d) 40cm3 of 0.2 mol.dm-3 hydrochloric acid.

e) 300cm3 of 4 mol.dm-3 sodium hydroxide solution.

5) Find the concentration in mol.dm-3 of these solutions:-

In question 3 you have to find the concentration of a solution starting from a mass of solute. The first step is to use the relative formula mass (RMM or Mr) to convert the mass to moles. (Remember that number of moles = mass in grams ( formula mass) Then continue as in question 1 above. Part a) is worked out for you as an example.

a) 20g of sodium hydroxide (NaOH) in 2dm3 of solution. (RAMs: Na = 23, O = 16, H = 1)

Ans:1 mole of NaOH = 23 + 16 + 1 = 40g : 20g ( 40g = 0.5 mole

( there is 0.5 mole of NaOH in 2dm3

( concentration = moles ( volume = 0.5 ( 2 = 0.25 mol.dm-3b) 10.6g of sodium carbonate (Na2CO3) in 1dm3 of solution. (RAM of C = 12)

c) 0.98g of sulphuric acid (H2SO4) in 200cm3 of solution. (RAM of S = 32)

d) 117g of sodium chloride (NaCl) in 5dm3 of solution. (RAM of Cl = 35.5)

e) 0.83g of potassium iodide (KI) in 25cm3 of solution. (RAMs: K = 39, I = 127)

6) Calculate the mass of solute in the following solutions:-

In question 4 you have to work out the mass of solute contained in the solution. First find the number of moles of solute, as in question 2 above, then convert the moles to a mass. Part a) has been done as a worked example for you.

a) 0.25dm3 of 2 mol.dm-3 calcium chloride solution. (RAM of Ca = 40)

Ans:moles = concentration ( volume = 2 ( 0.25 = 0.5 mole

1 mole of CaCl2 = 40 + (35.5 ( 2) = 111g

( 0.5 mole = 0.5 ( 111g = 55.5g of solute

b) 2dm3 of 0.2 mol.dm-3 potassium hydroxide (KOH) solution.

c) 200cm3 of 0.1 mol.dm-3 sodium carbonate (Na2CO3) solution.

d) 25cm3 of 0.05 mol.dm-3 copper (II) sulphate (CuSO4) solution. (RAM of Cu = 64)

e) 1.5dm3 of 0.4 mol.dm-3 nitric acid (HNO3). (RAM of N = 14)

Concentration of solution=number of moles of solute

(moles per dm3)volume of solution in dm3

moles of solute

concentration (mol.dm-3)

volume of solution (dm3)

1