How much heat is absorbed when 1 kg of a liquid gets vaporized at its boiling point

22.6 x 10^5 J

‘Heat of vaporization’ is the amount of ‘heat’ required to convert '1g' of mass of a liquid to vapor. Normal “boiling point of water” is 100 C. The heat of vaporization at this temperature is 2260 J/g. It means 2260 J/g of heat is required to convert ‘1g of water’ to 1g of vapor at 100 C. So to convert 1 kg of mass of water or 1000g of water we require 22.6 x 10^5 J of heat.

☆ionisation energy increases across a period from na -> ar☆ionisation energy decreases down a group from li -> fr☆ionic radius decreases across a period from na -> cl☆ionic radius increases down a groupfrom f -> ii hope it clarifies your doubts