Question

e) Suggest two ways of increasing the concentration of an aqueous solution. Concentrate please! A student had a solution of sodium chloride with a concentration of 93.6g/dm'. Calculate the mass of sodium chloride dissolved in 25.0cm' of the solution Extension questions: g) Concentration can be expressed in units of mol/dm'. A student has a solution of 73g/dm HCl. What is the concentration of the hydrochloric acid in mol/dm? (A H = 1, CI = 35.5) h) A student has a solution of 0.4mol/dm NaOH..What is the concentration of the hydrochloric acid in g/dm'? (A. Na = 23,0 = 16, H= 1) 1) A student dissolves 5.85g of sodium chloride in 250cm. Calculate the concentration of the sodium chloride in g/dm and mol/dm. (A. Na = 23, CI = 35.5)

Answer 1

e For increasing the concentration of an aqueous Solutions c) Reduce the volume of solvent ie water by evaporating. (01) Ü) Increase the amount of Solute by adding more Solute to the solvent. Concentration of Sodium chloride Solution = 93.69), 3 which means I dm3 (decimetre 3) soliition contains 93.6 grams of Sodium Chloride. 2(dm)* = 1 Cio'm) = 2(16.10?cm)) = Cio'cm) 3 = 10 cm ... 1 dm² = 103 cm :: Idma-hõcm? solution Contains > 93.6 9. of Nach 25.0 cm? solutions Contains, ? The amount of Nad = [(25.0).ems x (93.687 1 (cm 1. = (2300) 8. = 2.349 the amount of Nad present in 25.0 cm Solution = 2.349

g) Concentration of Hel solution with the student = 738/dm3 To convert the amount of Hcl from grams to moles, this amount should be divided with molar. mass of Ha. Molecular mass of Hel = ſixcatena. pras) + 14 ( =[(x1) + 1x(35.5) – 36.5 amu ... Molar mass of Hd = 36.5 g/mol. ma Lic mass Ekt Atomic I massa cl A ... Number of moles of Hel= (Mass of Hel (Molarmass of Hef (739 - =2 moles. (36.5g/mol) of Hul solution - 2 moles/dm3 :: Concentration Concentration = 2 mol/dm3 (b) Concentration of Naot solution = 0.4 mol/dm3 Molar mass of NaOH = 40 g/mol. T:. Molecular mass 1. Mass/Amount of Nacht Lo Nach = f Number of moles) xcmolarmass) = (Tx(Na)+ l< (OJ+1xCH FX1 X23) #{1*16)+(x1)} =(4 mol.) x 40(8mol) - 23+1671) = 16 g = 40 amu Concentration of Nach = 16 g/dm3

(sclute) (L) Amount of Naci = 5.859 Volume of solution - 250 cm3 cmd= 1 (16?m) = 1 (10! 16m)= (10'dm) = 1ődmº Zero= 1ődm? : Volume of solution = 250 cm) = (250 x 1534m3 = 0.25 dm3 Concentrations in g/4m? – (5.859 ) 10.25 dm3) - 23.4 g/dm3 * Number of moles of Nad = (mass of Nad . (Mdarmass of Nacy) : Molarmass of Nadi = 58.5g/ml. Mass of Nad = 5.859 :: Moleciclar mass of Nad = fix(Na) +l«CC) } Number of moles (n) = (Mass 1 6.5 (Malarmans) = [1 X23)+(x35:5) 15858 = 58.5 amy (58.5 g/mol) = 0.1 mol ::. Concentrations in mol/dm3 = (0molt 10.25 dm3 = 0.4 mol/dms