Question

Amount in gram of sample containing 80% NaOH required to prepare 60 litre of 0.5 M solution is

Answer 1

1 litre 1 M NaOH = 40 gm NaOH
Hence, 60 litre 0.5 M NaOH = (40×60) ÷ 2 gm
= 1200 gm
Required amount is = (1200 ÷ 80) × 100 gm
= 1500 gm (Ans)

Answer 2

Hello Dear.

Given that, Volume of the solution = 60 Litre .
and the molarity = 0.5 M.

Now we know the formula,
Molarity = Number of moles of solute / Volume of solution.
Hence Number of moles of solute = Molarity x Volume of solution.

Hence Number of Moles of NaOH = 60 x 0.5
=60 moles

Now , Molar mass of NaOH = 40 g/mole.
∴ Mass = Number of moles x Molar mass of NaOH.
 Mass = 30 x 40 =1200 g.

Hence Mass of NaOH = 1200 g.


Let the mass of the solution be 'm' gram.

then , Mass of NaOH = (80/100) x m 
 1200 = (80/100) x m .
  m=  1500 g.
Hence mass of the solution = 1500 g.


Hope it Helps.