પ્રથમ પધ્ધતિ મોલારીટી $ = \,\,\frac{{weight \,\,of\,\,soule\,\, \times \,\,1000}}{{Molecular\,\,\,Weight\,\,of\,\,solute\,\, \times \,\,volume\,\,of\,\,solution\,\,\left( {mL} \right)}}$
$ = \,\,\frac{{0.4}}{{40\,\, \times \,\,40}}\,\, \times \,\,1000\,\, = \,\,0.25\,\,M$
અને $normality\,\, = \,\,\frac{{weight\,\,of\,\,solute}}{{Equivalent\,\,weight\,\,of\,dolute\,\, \times \,\,volume\,\,of\,\,solution\,\,\left( {ml.} \right)}}\,\,$ $ \times \,\,1000$
$NaOH$ નો તુલ્યભાર $=\,40$
તેથી $, N\,\, = \,\,\frac{{0.4}}{{40\,\, \times \,\,40}}\,\, \times \,\,1000\,\, = \,\,0.25\,$
દ્રિતય પધ્ધતિ $N\,\, = \,\,M\,\, \times \,\,n $
અહી $n\,\, = \,\,1,\,\,\,\,N\,\, = \,\,M,\,\,\,N\,\, = \,\,0.25$
$\mathrm{HX}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{X}(\mathrm{aq}), \mathrm{Ka}=1.2 \times 10^{-5}$
$\left[\mathrm{K}_{\mathrm{n}}:\right.$ વિયોજન અચળાંક]
$300 \mathrm{~K}$ પર $HX$ ના $0.03 \mathrm{M}$ જલીય દ્રાવણ નું અભિસરણ (પરાસરણ) દબાણ .......... $\times 10^{-2}$ bar છે.
$\left[\right.$ આપેલ : $\mathrm{R}=0.083 \mathrm{~L} \mathrm{bar} \mathrm{Mol}^{-1} \mathrm{~K}^{-1}$ ]