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The mole fraction of B in vapour phase which is in equilibrium with this solution is .
0.16
5 molal: 5 mol B in 1000 g A (=20 mol A), so xB= 525=0.2, xA=0.8
P=xApA +xBpB 100=0.8 105+0.2 pB pB =80 mm Hg
Dalton's law: yB= xBpB P= 0.2 80100=0.16
Answer: 0.16
At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm-3) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm-3, the molar mass of the macromolecule is calculated to be X 104 g mol-1. The value of X is . Use: Universal gas constant (R) = 8.3 J K-1 mol-1 and acceleration due to gravity (g) = 10 m s-2
Vessel-1 contains w2 g of a non-volatile solute X dissolved in w1 g of water. Vessel-2 contains w2 g of another non-volatile solute Y dissolved in w1 g of water. Both the vessels are at the same temperature and pressure. The molar mass of X is 80% of that of Y. The van’t Hoff factor for X is 1.2 times of that of Y for their respective concentrations. The elevation of boiling point for solution in Vessel-1 is % of the solution in Vessel-2.