Question

The vapour pressure of benzene is 200mm . lowering of vapour pressure of 0.2molal solution is

Answer 1

Answer:

Explanation:

We know πV = nRT

πV = wB/mB RT

mB = wB×RT/πV     …(i)  

where wB = mass of solute (21.6× 10−3 g)

mB = molar mass of solute  

R = 0.0821 L atm K−1 mol−1

T = 298 K

V = 100/1000 = 0.1 L : π = 3.70/760 atm  

From (i), mB = 21.6×10–3×0.0821×298/(3.70/760)×0.1 = 1085