{:(,"Column I",,"Column II",),((A),"In 1 Maqueous NaoH, weight % of NaOH is " (d_("solution") = 2 gm//ml),(P),20,),((B),"Molarity of '22.4 V' " H_(2)O_(2) " solution",(Q),10,),((C ),"Molality of 20 ppm aqueous NaOH solution",(R ),3,),((D),"Mass % of " SO_(3) "in 102.25 % oleum sample",(S),2,),(,,(T),5 xx 10^(-4),):}

{:(,"Column I",,"Column II",),((A),"In 1 Maqueous NaoH, weight % of Na

1.	{:(,"Column I",,"Column II",),((A),"In 1 Maqueous NaoH, weight % of NaOH is " (d_("solution") = 2 gm//ml),(P),20,),((B),"Molarity of '22.4 V' " H_(2)O_(2) " solution",(Q),10,),((C ),"Molality of 20 ppm aqueous NaOH solution",(R ),3,),((D),"Mass % of " SO_(3) "in 102.25 % oleum sample",(S),2,),(,,(T),5 xx 10^(-4),):}
Answer» Solution :(A) 1 M aq. NaOH solution Mass of solute = 40 gm Mass of solution `= (1000 xx 2)` = 2000 gm `% (w//w) = (4)/(2000) xx 100 = 2%` (B) `M = (V)/(11.2) rArr (22.4)/(11.2) = 2M` (C ) 20 ppm NaOH (aq) solution mass of solution = 20 gm Mass of solution `= 10^(6) gm ~=` mass of solvent Molality `= ((20//40))/(10^(6)) xx 1000 = (1)/(2) xx 10^(-3)` `rArr 0.5 xx 10^(-3)` `rArr 5 xx 10^(-4)` (D) 102.25% Oleum sample 2.25 gm of `H_(2)O` `{:(SO_(3),+,H_(2)O,rarr,H_(2)SO_(4),),((1)/(8)mol,,(2.25)/(18) = (1)/(8) mol,,,):}` % mass of `SO_(3) = ((1)/(8) xx 80)/(100) xx 100 = 10%`