Question

1)  Convert in to
mole:

a)  70 g of Co2

b) 100 g of KOH

c) 50 g NaCl

d) 0.5 g of HCl


2) What is the mass of,

a) 10.5 u of Oxygen gas

b) 22.5 u  of water


3) How many molecules present in 70 g of Chlorine gas?


4) How many particules present in 140 g of Hydrogen gas?


5) Write the atomicity of the following:

a)Potassium

b) Neon

c) KMnO4

d) Ca(OH)2

Answer 1

1)  
       70 gms of CO2  --    molecular mass = 44   
               number of moles = 70/44 
     100 gms of KOH  --  molecular mass = 39+16+1 = 56.    number of moles = 100/56

       50 gm Na Cl   --    molecular mass = 23 + 35.45 = 58.45 
                         number of moles = 50/58.45 moles
        0.5 gms of H Cl   --  molecular mass = 1+35.45 = 36.45 
                   number of moles = 0.5/36.45

2)    10.5 u of oxygen gas --  1 u = 1.66 * 10⁻²⁴ gms   so multiply with 10.5
          22.5 u of water  -  weight / mass is  22.5 * 1.66 * 10⁻²⁴ gm

3) 70 g of chlorine gas -  molecular weight of Cl2 molecule is 70 u.  So 70 gm of chlorine gas is equal to one mole of chlorine gas. One mole of chlorine gas contains : N ( Avogadro number) of molecules. N = 6.023  * 10²³.

4) 140 gms of Hydrogen gas.  In gaseous form, Hydrogen is in H2 form. So molecular weight is 2. Hence molar mass is 2 gm per mole.  Hence there are 70 moles of hydrogen molecules.  Hence there are 70 * N number of molecules.  N = Avogadro number.

5) atomicity is the number of atoms in a molecule of a compound.
    a) potassium - K -  one
   b)  Neon  -   Ne  -  one
    c)  K Mn O4  - six atoms.  so six
    d)  Ca (OH)2   --    5 atoms  - so  5.