1.

If 1.71 g of sugar (molar mass=342) is dissoved in 500 cm^(3) of a solution at 300K. What will be its osmotic pressure?

Answer»

Solution :MASS of solute=`W_(2)=1.71g"Molar mass of solute"=M_(2)=342`
Volume of solution=500`CM^(3)=0.5L`Temperature=300K
R=0.083 L bar `mol^(-1)K^(-1)`
Formula: `pi=(W_(2)RT)/(M_(2)V)`
`pi=(1.71xx0.083xx300)/(342xx0.5)`
answer `pi=0.249` bar.


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