8.0575/100 KG OF GLAUBERS SALT. SALT IS DISSOLVED IN WATER TO OBTAIN 1dm(cube) OF A SOLUTION OF DENSITY 1077.2kg/m(CUBE) . CALCULATE THE MOLARITY ,MOLALITY AND MOLE FRACTION OF SODIUM SULPHATE IN THE SOLUTION....

as glauber's salt is Na₂SO₄ . 10H₂O

M of soln= d*v 

 M of soln -8.0475/100 = mass of water in d original solution.

but v also have to add contribution of 10H2O from salt  so mol wt of glauber's salt gives 10*18 g water so -8.0475/100 gives ?  this ans shd b added to get wt of solvent in kg.

now M, m can b calculated. 

mass of GS = 80.575 g

molar mass of GS = 322g

volm of soln = 1000mL

hence molarity = 0.25 M

mass of soln = 1077.2 kg/m3 x 1L = 1077.2 g

so mass of solvent = 1016.7 g

molality =0.24 m

322g of GS has 180g of water

so 80.575 g of GS has 45.122 g of water

now mass of Na2SO4 = 35.443 g

hence no of moles of Na2SO4 = 35.443/142 = 0.25

no of moles of H2O = 1016.7/18 =56.48

mole fraction of Na2SO4 = 0.25 / (0.25+ 56.48) = 0.0043

:)

its  IIT-94 prob...chkd d ques...

it says find the molarity n molality of glaubers salt n mole fractn of Na2SO4 in d soln