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Q.Define the following terms -

(a) Normality
(b) Molarity
(c) Molar fraction
(OR)
4 gm of caustic soda (molar mass 40) is dissolved in water to prepare 200 gm of solution. Calculate the molality of the solution and molar fraction of the solute.
Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2023Subjective· 3mImportance★★★★★
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Three ways of expressing concentration: normality (equivalents/L), molarity (moles/L), mole fraction (moles of one component / total moles).

  1. Normality (N): The number of gram equivalents of solute dissolved per litre of solution. N=No. of gram equivalents of soluteVolume of solution (in litres)N = \dfrac{\text{No. of gram equivalents of solute}}{\text{Volume of solution (in litres)}}
  2. Molarity (M): The number of moles of solute dissolved per litre of solution. M=No. of moles of soluteVolume of solution (in litres)M = \dfrac{\text{No. of moles of solute}}{\text{Volume of solution (in litres)}} (c) Molar fraction (mole fraction, xx): The ratio of the number of moles of a particular component to the total number of moles of all components present in the solution. xA=nAnA+nBx_A = \dfrac{n_A}{n_A + n_B} OR (numerical problem): 4 g of caustic soda (NaOH, M=40 g/molM=40\ g/mol) dissolved to make 200 g of solution. Moles of NaOH (solute) =440=0.1 mol= \dfrac{4}{40} = 0.1\ mol Mass of water (solvent) =200−4=196 g=0.196 kg= 200 - 4 = 196\ g = 0.196\ kg Molality =0.1 mol0.196 kg=0.510 mol/kg (≈0.51 m)= \dfrac{0.1\ mol}{0.196\ kg} = 0.510\ mol/kg\ (\approx 0.51\ m) …

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