The boiling point of `0.1 mK_(4)[Fe(CN)_6]` is expected to be `(K_(b)` for water = `0.52K kg mol^(-1)`)

The boiling point of 0.1 molal K_4 [Fe(CN)_6 ] solution will be (given K_b for water = 0.52 k kg mol^(-1))

The boiling point of 0.1 molal K_(4)[Fe(CN)_(6)] Solution will be ( given Kb for water "=0.52K" )

A solution of urea has boiling point 100.15^@C, K_f and K_b for water are 1.87 and 0.52 K kg "mol"^(-1) . Calculate (a) molality and (b) freezing point of the solution.

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)

A solution contains 5.6 gm of glucose in 1000 gm of water. Calculate the boiling point of the solution. ( k_b for water = 0.52 K kg mol^-1 )

Elevation in boiling point of an aqueous glucose solution is 0.6, K_b for water is 0.52 K molal ""^(-1). The mole fraction of glucose in the solution is:

Calculate the boiling point of solution when 4g of Mg SO_(4) (M=120g "mol"^(-1)) was dissolved in 100g of water, assuming MgSO_(4) undergoes complete ionization (K_(b) " for water " = 0.52 K " kg mol"^(-1))

0.52g of glucose (C_6H_12O_6) is dissolved in 80.2g of water calculate the boiling point of the solution. ( K_b for water is 0.52 K kg mol^-1 ).

The freezing point (in .^(@)C) of a solution containing 0.1g of K_(3)[Fe(CN)_(6)] (Mol. wt. 329 ) in 100 g of water (K_(f) = 1.86 K kg mol^(-1)) is :