please provide solution to my query experts.

Dear Student
mass of ethyl alcohol (C₂H₅OH) = 23g
molar mass of C₂H₅OH = 46 g/mol
moles of C₂H₅OH = 23/46 = 0.5 mol

Volume of water = 100 ml = 0.1 L
density of solution = 0.5 g/ml

(i) $molarity\left(M\right)=\frac{moles of solute}{Volume of solution\left(L\right)}$
                          = 0.5/ 0.1 = 5M
(ii) Volume of water = 100 ml [Volume of water = Volume of Solution] 
density of solution = 0.5 g/ml
mass of solution = 0.5 x 100 = 50g
mass of ethyl alcohol = 23g
mass of water (solvent) = (50 -23)g = 27g

$molality\left(m\right) = \frac{moles of solute}{weight of solvent \left(kg\right)}$
                       = 0.5/(27/1000)
                       = 18.51m
(iii) moles of solute (ethyl alcohol) = 0.5 mol
      moles of solvent (water) = 27/18 = 1.5 mol

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$\chi = \frac{moles of solute}{moles of solute + moles of solvent}$
    = 0.5/(0.5 + 1.5) = 0.5/2 =0.25

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Regards