When `0.6g` of urea dissolved in `100g` of water, the water will boil at `(K_(b)` for water `= 0.52kJ. mol^(-1)` and normal boiling point of water `=100^(@)C)`:

When 0.6 g of urea is dissolved in 100 g of water, the water will boil at ( K_(b) for water = 0.52 Km^(-1) and normal boiling point of water = 100^(@)C ) :

When 0.6 g of urea dissolved in 100 g of water water will boil at ( K_(b) for water 0.52kJ"mol"^(-1) and normal boiling point of water =100^(@)C )

When 0.6 g of urea is dissolved in 100 g water, the water will boil at (Kb, for water = 0.52 K/m and normal boiling point of water = 100^@C ):

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) )

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)

18 g of glucose is dissolved in 1 kg of water. At what temperature will the solution boil ? ( K_b for water is 0.52 K kg "mol"^(-1) )