100ml of ethyl alchoal [d=0.92g/ml] and 900ml of water [d=1g/ml] are mixed to form 1lit soln the molarity and molality of soloution is

Given: 100ml of Ethyl alcohol and 900 ml of water
Mass of ethyl alcohol present: Mass= Density*Volume= 100*0.92=92g
Mass of water present= 900g.

No. of moles of ethyl alcohol; Given mass/molecular mass of ethyl alcohol=92/46= 2 moles

Molarity= No. of moles of solute/1 litre of solution

=2/1 = 2M


Therefore molarity= 2M.

Molality = No. of moles of solute in 1k of Solvent
In 900 g of water, 92 g of ethyl alcohol is present.
In 1g of water, 92/900 g of ethyl alcohol is present. 
Therefore, in 1000 g of water, (92/900) * 1000g of ethyl alcohol is present.
= 102.22 g
Therefore number of moles = 102.22/46
=2.22 ( approx )

therefore molality = 2.2m