The creysopic constant `(K_(f))` for water is 1.86 K `mol^(-1)kg^(-1)`. What does it signift ?

The cryoscopic constant of water is 1.86 K "mol"^(-1) kg. An aqueous solution of cane sugar freezes at -0.372^(@)C . Calculate the molality of the solution.

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

What do you understand by the term that K_(f) for water is 1.86 K kg mol^(-1) ?

Depression in freezing point is 6K for NaCl solution. If K_(f) for water is 1.86 "K kg mol"^(-1) , amount of NaCl dissolved in 1kg water is :

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

The freezing point of a 1.00 m aqueous solution of HF is found to be -1.91^(@)C . The freezing point constant of water, K_(f) , is 1.86 K kg mol^(-1) . The percentage dissociation of HF at this concentration is:-

When a solution containing "W" g of urea in 1 kg of water is cooled to -372(@)C , 200g of ice is separated . If K_(f) for water is 1.86 K kg "mol"^(-1) . W is ___________