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# Concentration of Solutions - Introduction

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Introduction

oncentration of Solutions - Introduction Some measure of the amount of solute dissolved in a solvent is needed when dealing with solutions (particularly aqueous solutions). e.fg. mass per unit volume : g cm-3, kg m-3 or g dm-3 amount per unit volume : mol m-3 or mol dm-3 N.B.: 1 dm3 = 1000 cm3 The amount concentration equation given above can be manipulated to give another useful equation : This equation must be used when dealing with volumes and solutions and must not be confused with another equation giving the amount of substance : back to top Concentration of Solutions - Calculations (1) Question : Sulphuric acid of concentration 0.1 mol dm-3 reacts exactly with 10 cm3 of 0.2 mol dm-3 sodium carbonate(aq). What volume of sulphuric acid is needed? ...read more.

Middle

Therefore, the amount of A used = Va � Ca Therefore, the amount of B used = Vb � Cb Since na mol of A reacts with nb mol of B these two equations may be combined to give, nb � Va � Ca = na � Vb � Cb In an examination question you will be given five out of the six pieces of information above and by rearrangement of the mathematical expression the sixth value can be worked out. e.g. in the above worked example between sulphuric acid and sodium carbonate, na = 1 nb = 1 Va = unknown Vb = 10 cm3 Ca = 0.1 mol dm-3 Cb = 0.2 mol dm-3 => Va = 20 cm3 or 0.02 dm3 back to top Titration N.B. ...read more.

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