Q1 .A sample of 0.025 mol of the chloride of an element Z was dissolved in distilled water and the solution made up to 500cm3. 12.5cm3 of this solution reacted with 25cm3 of 0.1 moldm-3 silver nitrate solution. What is the formula of the chloride?


A. Z2Cl    B. ZCL   C. ZCl2   D. ZCl4


Help:

Silver (I) Chloride reacts with Chloride to give silver (I) chloride percipitate : 

                                Ag+ (aq)    +   Cl- (aq) -------------> AgCl (s)

The amount of Aq+ used? = Concentration X volume

                                                       = o.1 (mol dm^-3)  x  (25/1000) (dm^3)
                                                      = 2.5 x 10^-3

Since 1 mole of Ag+ react with 1 mole of Cl- ,  Amount of Cl- present in 12.5 cm^3     =2.5x10^-3/12.5 x 500
                                                                                                                                                                             = 0.1 mol


Since 0.025 mole of compound contain o.1 mole of chloride, I mole of the compound therefore contain
0.1/0.025 = 4 mole of chloride. 

The Formula for the compound is ZCl4

Answer is D

Q2)  In the absence of a catalyst, ammonia burns in an excess of Oxygen to produce steam and nitrogen. What is the volume of oxygen, remaining when 60 cm^3 of ammonia is burnt in 100 cm^3 of oxygen, all  the volume being measured at the same temperature and pressure?

A 25 cm^3   B 35cm^3  C 40 cm^3  D 45 cm^3  E 55cm^3



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4 NH3 (g)  +  3O2 (g)  ----------> 2N2 (g) + 6 H2O (g)


60 cm^3 of NH3  requires 3/4 X 60 = 45cm^3 of O2

Therefore, Volume of Excess O2= 100-45=55 cm^3



Q4 A sample of 10dm^3 of populated air is passed through the lime water so that all of the carbon dioxide present is precipitated as calcium carbonate. The mass of calcium carbonate formed is  0.05 g. What is the percentage, By volume of  carbon dioxide in the air sample?

{Relative Atomic Masses C, 12 O 16; Ca 40; I mole of gas under the experimental conditions has a voulme of 24 dm^3}


A 0.03%  B  0.05  %   C 0.12%  D 0.3 %


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Ca(OH)2 (aq)  +  CO2(g)-----------> CaCO3 (s) + H2O

Amount of precipitated = 0.05/40+12+3 x 6
                                                 =  5.0 X 10^-4 x 24
                               
Therefore the volume of CO2= 5.0  x 10^-4 x 24
                                                = 1.2x10^-2  dm^3

Therefore, % volume of CO2 =1.2 x 10^-2 /  10    X 100
                                                = 0.12

Answer C


Q5

A sample of 2.00 g of iron (III) sulphate, Fe2(SO4)3, is dissolved in water to give 100 cm^3 of aquous solution. What is the concentration of SO4^2- ions? { The relative formula (molecular) mass of Fe2(SO4)3 IS 400} 


A 1.5 x 10^-3 mol dm^-3
B   5  x 10^-3 mol dm^-3
C 1.5 x 10^-2 mol dm^-3
D 5    x 10^-2 mol dm^-3
E 1.5 x 10^-1 mol dm^-3


Help

1 mole of Fe2 (So4)3 gives 3 mole of SO4^2-  ions.
Therefore, Concentration of SO4^2- = (2.00/400  /   100/1000)   X  3
                                                           = 0.15 mol dm^-3


Q6

What is the volume of Oxygen is required for the complete combustion of mixture of 5 cm^3 of CH4 and 5 cm^3 of C2H4?

A 5cm^3   B 10 cm^3  C  15 cm^3  D 20 cm^3 E 25 cm^3

Help

CH4 (g) + 2O2 (g)--------> CO2(g) + 2H2O (l)
C2H4 (g) + 3O2(g)------->2CO2 (g) + 2H20 (l)

Volume of O2 required to burn 5 cm^3 of CH
= 2x5=10cm^3
Volume of O2 required to burn 5 cm^3 of C2H4
= 3 x 5 =15 cm3
Total Volume of O2 required
= 10+15=25cm^3

Answer E



Q7 In which of the following reactions does hydrogen behave as an oxidizing agent?

A H2 + Cl2---------> 2HCl
B C2H4 + H2------->C2H6
C C2H5CHO + H2------->C2H5CH2OH
D N2 + 3H2-------->2NH3
E 2Na + 3H2-------->2NaH



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An Oxidising