In an experiment 1g of nonvolatile solute was dissolved in 100g of acetone(mol.mass-58)at 298k.The vapour pressure of the solution was found to be 192.5mm Hg.The molecular weight of the solute is (vapour pressure of acetone=195mm Hg)
Answers
The answer is 45.37
Explanation:
Mass of non-volatile solute, m 1 = 1 g
Mass of solvent (acetone), m2 = 100 g
Molar mass of solvent, M2 = 58.08 g
T = 298 K
Vapour pressure of pure ethyl acetate, p2 o = 195mm Hg [it cannot be 1.95, it has to be 195mmHg
Vapour pressure of solution, ps = 192.5mm Hg
Mole fraction of the solute can be determined by using the formula for relative lowering of vapour pressure
( p20−psp20)/p20=(m1×M2M1×m2)
)195−192.5)/195=(1×58.08)/(M1×100)
0.0128 =(58.08)/(100×M1)
0.0128M1 =0.5808
M1 =45.37g
The molecular weight of the solute is 45.37 g.
Given:
Molar mass of acetone = 58 g/mol = M2
Mass of acetone = 100 g = m2
Mass of solute = 1 g = m1
Vapor pressure of the solution = 192.5 mm Hg = p1
Vapor pressure of acetone = 195 mm Hg = p2
Explanation:
The relative lowering of the pressure is given by the formula:
(p2 - p1)/p2 = (m1 × M2)/(m2 × M1)
On substituting the values, we get,
(195 − 192.5)/195 = (1 × 58.08)/(100 × M1)
2.5/195 = (58.08)/(100 × M1)
0.0128 = (58.08)/(100 × M1)
100 × M1 = (58.08)/0.0128
100 × M1 = 4537.5
M1 = 4537.5/100
∴ M1 = 45.37 g