A solution containing`0.5g` of a non-volatile solute in `0.2dm^(3)` of the solution exerts an osmotic pressure of `44.44 kPa` at `300K`. Thus, molar mass of the solute is

Vapour pressure of a solution containing a non volatile solute is

A solution containing 30g of non-volatile solute exactly in 90g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: molar mass of the solute

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: molar mass of the solute

3 gram of non-volatile solute in a 1000 cm^(3) of water shows an osmotic pressure of 2 bar at 300K. Calculate the molar mass of the solute (R=0.0853" L bar "K^(-1)mol^(-1)) .

A solution containing 30 g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate: vapour pressure of water at 298 K.

A solution is made by dissolving 30 g of a non-volatile solute in 90 g of water. It has a vapour pressure of 2.8 k Pa at 298 K. At 298 K, vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of the solute.

3.0 g of non-volatile solute when dissolve in 1 litre water, shows an osmotic pressure of 2 atmosphere at 300 K. Calculate the molar mass of the solute. (R = 0.0821 litre atm K^(-1)"mol"^(-1) ).

A solution containing 30g of a non-volatile non-electrolyte solute exactly in 90g water has a vapour pressure of 2.8 kPa at 298K. Further, 18g of water is then added to solution, the new vapour pressure becomes 2.9kPa at 298K. The solutions obey Raoult's law and are not dilute, molar mass of solute is