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Question

# (IE)1 and (IE) 2 of Mg are 740,1540kJ mol-' calculate the percentage of Mg+ and Mg2+ if 1g of Mg absorbs 50.0KJ of energy.

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Solution

## Let x be the mole of Mg+ Let y be the mole of Mg2+ A total mole of Mg in 1 gm of Mg = 1/molar mass of Mg = 1/24So x + y = 1/24Energy absorbed by 1 mole of Mg+ = 740 KJSo energy absorbed by x mole of Mg+ = 740x KJEnergy absorbed by 1 mole of Mg2+ = 1540 KJSo enegy absorbed by y mole of Mg2+ = 1540y KJTotal energy absorbed = 50 KJ So 740x KJ + 1540y KJ = 50 KJ 740x + 1540y = 5074x + 154y = 5x + y = 1/2424x+24y = 174x + 154y = 5Substitute the value of x from equation 1 to equation 2x = (1/24 - y)74 * (1/24 - y) + 154y = 5154y - 74 y = 5 - 74/2480y = (120-74)/2480y = 46/24y = 23/960 moleMass of Mg2+ present = 23/960 * 24 = 23/40 gm So x = (1/24 -y)x = 1/24 - 23/960x = (40-23) /960 x = 17/960 moleMass of Mg+ present = 24 * 17/960 = 17/40 gm Percentage of x = x/(x+y) *100 = 17/40 * 100 = 42.5 %Percentage of y = 100 - 42.5 = 57.5 %Answer : Mg+ = 42.5 % and Mg 2+ = 57.5 %

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