Question

An aqueous solution of a soluble compound (a nonelectrolyte) is prepared by dissolving 33.2 g of the compound in sufficient water to form 250 ml of solution. the solution has an osmotic pressure of 1.2 atm at 25 °c. what is the molar mass (g/mole) of the compound?

Answer 1

M=n(pie)/RT
n=osmotic pressure(1.2 atm)
M=molar  of the  solution
R=gas  constant(0.0821)
T=  temperature  in  kelvin 25+273
M=[1.2atm /(0.0821L atm/k mol x  298k)]=0.049mol L
M= moles  of  the  solute/ litres  of solution(250/1000)
0.049=  y/0.25
moles  of  solute is therefore =0.01225mol
molar  mass=33.29 g/0.01225mol=2.7 x10^3g/mol

Answer 2

Answer : The molar mass (g/mole) of the compound is, 2707.55 g/mole

Solution : Given,

Mass of solute = 33.2 g

Volume of solution = 250 ml = 0.250 L

Temperature of solution = $25^oC=273+25=298K$

Formula used for osmotic pressure :

$\pi=\frac{nRT}{V}\\\\\pi=\frac{wRT}{MV}$

where,

$\pi$ = osmotic pressure

V = volume of solution

R = solution constant  = 0.0821 L.atm/mole.K

T= temperature of solution

M = molar mass of solute

w = mass of solute

Now put all the given values in the above formula, we get the molar mass of the compound.

$1.2atm=\frac{(33.2g)\times (0.0821Latm/moleK)\times (298K)}{(M)\times (0.250L)}$

$M=2707.55g/mole$

Therefore, the molar mass (g/mole) of the compound is, 2707.55 g/mole