4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre ofwater. The molality of the resultant solution is _____________.(i) 0.004(ii) 0.008(iii) 0.012(iv) 0.016

EXPLAIN IN DETAIL.

Molarity of NaCl = 0.02 M
No. of moles of NaCl =0.02 x 4 = 0.08 moles

Final Volume of solution = 4 +1 =5 litres
Mass of 5 L water  = 5 kg

Molality = No. of moles of solute / kg of solvent = 0.08 moles / 5 kg = 0.016 m

  • 26

as M1V1=M2V2

thus we have 4*0.02=M2*(4+1)

M2=0.016

  • 2
What are you looking for?