The reduction reaction of the oxidising agent(MnO4–)
involves gain of 3 electrons.
Hence the equivalent mass = (Molar mass of KMnO4)/3
= 158.1/3 = 52.7
No. of moles of water present in 180 g
= Mass of water / Molar mass of water
= 180 g / 18 g mol-1 = 10 moles
One mole of water contains
= 6.022 x 1023 water molecules
10 mole of water contains = 6.022 x 1023
x 10 = 6.022 x 1024 water molecules
7.5 g of gas occupies a volume of 5.6 liters at 273
K and 1 atm pressure Therefore,
the mass of gas that occupies a volume of 22.4 liters
=7.5g / 5.6 L = 30g
Molar mass of NO (14+16) =
30 g
No. of electrons present in one ammonia (NH3)
molecule (7 + 3) = 10
No of molesof ammonia
=Mass/Molar mass
=1.7g/ 17 gmol-1
=0.1 mol
No of molecules present in 0.1 mol of ammonia
= 0.1x 6.022 x1025= 6.022 x 1022
No of elec . n trons present in mol of ammonia
=10 x 6.022 x1022 = 6.022x1023
No, of moles of oxygen=8 g/32 g
=0.25 moles of oxygen
No. of moles of sulphur dioxide=8 g / 64 g
=0.125 moles of sulphur dioxide
Ratio between the no. of molecules=0.25: 0.125
=2:1
AgNO3 + KCl → KNO3 + AgCl
50 mL of 8.5 % solution contains 4.25 g of AgNO3
No. of moles of AgNO3 present in 50 mL of 8.5 %
AgNO3 solution
=Mass / Molar mass
=4.25 / 170
=0.025 moles
Similarly, No of moles of KCl present in 100 mL of
1.865 % KCl solution
=1.865 / 74.5
=0.025 moles
So total amount of AgCl formed is 0.025 moles
(based on the stoichiometry)
Amount of
AgCl present in 0.025 moles of AgCl
=no. of moles x molar mass
=0.025 x
143.5 = 3.59 g