Home
Class 12
CHEMISTRY
3.795 g of sulphur is dissolved in 100 g...

3.795 g of sulphur is dissolved in 100 g of carbon disulfide. This solution boils at 319.81K. What is the molecular formula of sulphure in solution? The boiling point of the solvent is 319.45K.
(Given that Kb for CS2 = 2.42 K kg mol(-1) and atomic mass of S = 32)

Promotional Banner

Similar Questions

Explore conceptually related problems

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 depression of freezing point of a solvent (b.p:303 K) for a particular solution is 0.153 K. Calculate the molal elevation constant if the boiling point of the solution is 304.52 K .(K_(f) =1.68 K kg mol ^(-1))

The depression of freezing point of watar for a particular solution is 0.186 K. The boiling point of the same solution is _______. (K_(f)= 1.86K kg mol^(-1) and K_(b)= 0.512 K kg mol^(-1))

IF 10 g of solute was dissolved in 250 mL. of water and osmotic pressure of the solution was found to be 600 mm of Hg at 300 K, then molecular weight of the solute is _______. g mol ^(-1)

3.4 g of CaCl_2 is dissolved in 2.5 L of water at 300 K. What is the osmotic pressure of the solution? van't Hoff factor for CaCl_2 is 2.47. (Ca = 40, Cl =35.5)

At 298 K, 1000 cm^3 of a solution containing 4.34g of solute shows osmotic pressure of 2.55 atm., What is the molar mass of the solute.