Molality of NaOH is 1.25,percent by …
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Dr. Kamlapati Bhatt 8 years, 6 months ago
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 %
0Thank You