4.0 g of NaOH are present in one decilitre of solution . Calcute Mole fraction of NaOH Molality of solution Molarity of solution.

Calculation of molarity of the solution
Mass of NaOH =4g, Molar mass of NaoH=40 g `mol^(-1)`
Volume of solution=1 decilitre=0.1L
Molarity (M)`=("Mass of NaOH"//"Molar mass of NaOH")/("Volme of solution in litres")`
=`((4g)//(40g mol^(-1)))/((0.1 L))=1 molL^(-1)=1 M`
Colulation of molality of the solution
Volume o solution100 mL, `(100 mL)xx(1.038g//mL)=103.8 g`
Mass of sovent=(103.8-4)=99.8g=0.0998 kg
Molality of solution (m)`=("Mass of NaOH "//" Molar mass of NaOH")/("Mass of solvent in kg")`
`=((4g)//(40gmol^(-1)))/((0.0998 kg))=1.002 mol//kg=1.002 m`
Calculation of mole fraction of NaOH
No. of moles of NaOH`=("Mass of NaOH")/("Molar mass")=((4g))/((40 g mol^(-1)))=0.1 mol`
`"No. of moles of water"=("Mass of "H_(2)O)/("Molar mass")=((99.8g))/((18 g mol^(-1)))=5.54 `mol
`"Mole fraction of" NaOH=n_(N_(a)OH)/(N_(a)OH+N_(H_(2)O))=((0.1 mol))/((0.1 mol)+(5.54 mol))=0.018`