5 g of Na_(2) SO_(4) was dissolved in x g of H_(2)O. The change in froczing point was found to be 3.82^(@)C. If Na_(2)SO_(4) is 81.5% ioniscd, the value of x (K_(f)for water =1.86 ^(@)C kg mol ^(-1)) is approximately: (molar mass of S = 32 g mol ^(-1)) and that of Na=23 g mol ^(-1))

5 g of Na_(2) SO_(4) was dissolved in x g of H_(2)O. The change in fro

1.	5 g of Na_(2) SO_(4) was dissolved in x g of H_(2)O. The change in froczing point was found to be 3.82^(@)C. If Na_(2)SO_(4) is 81.5% ioniscd, the value of x (K_(f)for water =1.86 ^(@)C kg mol ^(-1)) is approximately: (molar mass of S = 32 g mol ^(-1)) and that of Na=23 g mol ^(-1))
Answer» 15 G 25 g 45 g 65 g Solution :Molarity (cxpcrimental) ` = (DELTA T _(f))/(K _(f)) = (3.82)/( 1.86) =(3.82)/(1.86)= 2.054` mol/1000 g solvent Molarity (thcartical) `= ("mole of solute")/("wt. of solventing(g))xx1000` `= ( 5g//142 g //"mole")/(x) xx1000` `Na_(2) SO_(4) to 2NA^(+) +SO_(4)^(2-)` `{:("Moles before dissociation", 1,0,0),("Moles after dissociation", 1-x, 2x, x):}` Von't FACTOR (i) `= ("Moles after dissociation")/("Moles before dissociation")` `= ((1-x 0 + 2x+x)/( 1)` `Na_(2) SO_(4)` is ionised `81.5%` means `x=0.815` `=((1-0.815) + 2xx0.815 +0.815)/(1) =2.63.` `i = ("Observed molarity ")/("Calculated molarity")` `implies 2.63 =(2.054)/((0.0352)/(x)XX 1000) =45.07g.`