Assming complete dissociation, Calculate the expected the expected freezing point of a solution prepared by dissolving 6.0 g of Glauber's salt. `Na_(2)SO_(4). 10H_(2)O` in 0.1 kg of water. (Given value of `K_(f)=1.86 K kg mol^(-1))`.

Define (i) Mole fraction (ii) Molality (iii) Raoult's law (b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber's salt, Na_(2)SO_(4).10 H_(2)O in 0.100 kg of water. ( K_(f) for water = 1.86 K kg "mol"^(-1) , Atomic massess : Na=23, S=32, O=16 , H=1)

Define (i) Mole fraction (ii) Molality (iii) Raoult's law (b) Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6.00 g of Glauber's salt, Na_(2)SO_(4).10 H_(2)O in 0.100 kg of water. ( K_(f) for water = 1.86 K kg "mol"^(-1) , Atomic massess : Na=23, S=32, O=16 , H=1)

Assuming complete dissociation, calculate the expected freezing point of a solution prepared by dissolving 6g of Na_2SO_4 .10H_2O~ in 0.1 kg of water ( k_f for water= 1.86)

What will be the freezing point ("in"^(@)C) of solution obtained by dissolving 0.1 g potassium ferricyanide (mol wt = 329) in 100 g water. If K_(f) for water is 1.86 K kg mol^(-1)

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol^(-1) ) in 250 g of water . ( K_(f) of water = 1.86 K kg mol^(-1) )

Calculate the freezing point of a solution containing 60 g glucose (Molar mass = 180 g mol^(-1) ) in 250 g of water . ( K_(f) of water = 1.86 K kg mol^(-1) )

Calculate the depression in the freezing point of a solution containing 0.1 g of K_(3)[Fe(CN)_(6)] in 100 g of H_(2)O ? (molecular weigth of K_(3)[Fe(CN)_(6)]=329, K_(f)=1.86K " kg mol"^(-1) )

The freezing point of a solution containing 0.1g of K_3[Fe(CN)_6] (Mol. Wt. 329) in 100 g of water ( K_f =1.86K kg mol^(−1) ) is:

A solution of sucrose (mol.mass=342) was prepared by dissolving 34.2g of it in 1000g water. Find its freezing point. (K_f for water = 1.86k kg mol^(-1) )