At certain Hill-station pure water boils at `99.725^(@)C`. If `K_(b)` for water is `0.513^(@)C kg mol^(-1)`, the boiling point of `0.69m` solution of urea will be:

Pure water boils at 99.725^(@)C at Shimla. If K_(b) for water is 0.51 K mol^(-1) kg the boiling point of 0.69 molal urea solution will be:

Pure water boils at 99.725^(@)C at Shimla. If K_(b) for water is 0.51 K mol^(-1) kg the boiling point of 0.69 molal urea solution will be:

Complete the following statements by selecting the correct alternative from the choices given : An aqueous solution of urea freezes at - 0.186^(@)C, K_(f) for water = 1.86 K kg. mol^(-1),K_(b) for water = 0.512 "K kg mol"^(-1) . The boiling point of urea solution will be :

Boiling point of water 750 mm Hg is 99.63^(@)C . How much sucrose is to be added to 500 g of water such that it boils at 100^(@)C . [K_(b) for water is 0.52" K kg mol"^(-1)] i) Since boiling point is changing, apply the formula for elevation in boiling point, Delta"T"_(b)=K_(b)m ii) m=(W_(B))/(M_(B).W_(A)) So, DeltaT_(b)=(K_(b).W_(B))/(M_(B)xxW_(A)) Or W_(B)=(DeltaT_(b)xxM_(B)xxW_(A))/(K_(b)) iii) Find DeltaT_(b)" as "DeltaT_(b)=T_(b)=T_(b)-T_(b)^(0) T_(b) = Boiling point of solution T_(b)^(0) = Boiling point of pure solvent

How much urea is to be added to 500g water so that it will boil at 374K? ( k_b for water = 0.52K kg mol^(-1) )

An aqueous solution freezes on 0.36^@C K_f and K_b for water are 1.8 and 0.52 k kg mol^-1 respectively then value of boiling point of solution as 1 atm pressure is

An aqueous solution freezes on 0.36^@C K_f and K_b for water are 1.8 and 0.52 k kg mol^-1 respectively then value of boiling point of solution as 1 atm pressure is

Calculate the boiling point of urea solution when 6 g of urea is dissolved in 200 g of water. ( K_(b) for water = 0·52 K kg mol^(-1) , boiling point of pure water = 373 K, mol.wt. of urea = 60)