5g of Nacl is dissolved in 1000g of water. If the density of the resulting solution is 0.997 g per "cc", calculate the molality, molality, normality and mole fraction of the solute.

5g of Nacl is dissolved in 1000g of water. If the density of the resul

1.	5g of Nacl is dissolved in 1000g of water. If the density of the resulting solution is 0.997 g per "cc", calculate the molality, molality, normality and mole fraction of the solute.
Answer» Solution :Mole of NaCl `=(5)/(58.5)=0.0854` (mol.wt.of `NaCl=58.5`) By definition : Molality `=("moles")/("wt.of solvent in grams")xx1000` `=(0.0854)/(1000)xx1000=0.0854m` VOLUME of the solution `=("wt in grams")/("density in grams"//"cc")=(1005)/(0.997)"cc"` `=1008mL=1.008` litres Again by definition molarity `=("moles")/("volume of solution in litres")=(0.0854)/(1.008)=0.085M` `:.` normality = `0.085N` (for NaCl, eq.wt.=mol.wt.) Further, `underset(("PER"1000g))("moles of" H_(2)O)=(1000)/(18)=55.55` (Supposing 1mL=1g for water having density `1g//mL`) TOTAL moles = moles of NaCl `+` moles of `H_(2)O` `=0.0854+55.55=55.6409` Mole fraction of NaCl=`("moles of" NaCl)/("total moles")=(0.0854)/(55.6409)` `=1.53xx10^(-3)`