1.

Aqueoussolutionof NaOH ismarked10%(w//w). Thedensityof thesolutionis 1.070 g cm^(-3). Calculate (i) molarity and (ii)molalityof NaOH.

Answer»


Solution :Given :Percentage byweightof NaOH`=10% (w//w)`
Densityof NaOHsolution =D =1.070 G `CM^(-3)`
Molarmass ofNaOH=40 g `mol^(-1)`
(i) Molarity of NaOHsolution = ?
(ii)Molalityof NaOHsolution = ?
(i) Formolarityof solution: Consider 100 gNaOHsolution
`:. ` Weight `H_(2)O+` Weightof NaOH =100 g
`:. ` Weightof `H_(2)O = 100 - 10= 90 g`
Numberof molesof NaOH`= (W_(NaOH))/(M_(NaOH))`
`:. n_(NaOH) = .(10)/(40)`
`=0.25 `mol
Densityof solution `(D) = ("Weight of solution" )/(" volumeof solution " ) = (W)/(V)`
`:.V = (W)/(D)`
`=(100)/(1.070)`
`=93 . 46 cm^(3)`
`=0.09346 dm^(3)`
Molarity`=(n)/(V)= (0.25)/(0.09346) =2.675 "mol"dm^(-3)(or M)`
(ii) For molalityof NaOHsolution :
Molality`=("MOLAR of NaOH ") /("Mass ofsolventin gram ") xx 1000`
`=(0.25)/(90) xx 1000`
`=2.779"mol" kg^(-1)(or m)`


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