Molality  of NaOH is 1.25,percent by …

step 1,

Molality (m)  =   No. of  moles / 1000 g of solvent ( water )

...................=  1.25 / 1000 g

.................. = ( 1.25 x40  ) g /  1000 g

.................. =  50 g /  1000 g 

{tex}\therefore{/tex} total weight of the solution =  1050 g

step 2 ,

Percent by weight of the solution  is given by the expression ,

......................................................................................( Mass of solute ) / [ ( mass of solute )  +  ( mass of solvent ) ]   

Since  , 1050 g of solution contains  solute ( NaOH )   =   50.00  g

{tex}\therefore{/tex}   100 g of solution would contain solutte ( ie. NaOH ) = ( 50 / 1050 )

.................................................................................= 0.0476

Thus  , the percent  by weight of NaOH in the solution  is  0.0476 %