Question

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

Answer 1

Hello Dear.

Volume of the solution = 60 liter.
Molarity of the solution = 0.5 M.

Using the Formula,
Molarity = No. of moles of solutes/Volume of the solutions in liter.
∴ 0.5 = No. of moles/60
⇒ No. of moles of NaOH = 30 moles.

Molar mass of the NaOH = 40 g/mole
[∵ 23 + 16 + 1 = 40]

∵ No. of mole = Mass/Molar Mass
∴ Mass = No. of moles × Molar Mass
⇒ Mass = 30 × 40
⇒ Mass = 1200 g.

∴ Mass of the NaOH = 1200 g.

Now, Let the Mass of the solution be x g.

According to the Question,

Mass of the NaOH = 80% of the Solution.
1200 = (80/100) × x
⇒ x = 15 × 100
∴  x = 1500 g



Hence, the mass of the sample or solution is 1500 g.


Hope it helps.

Answer 2

We know,
Molarity = percentage of solute × 10 × density of solution /molar mass of solute
Given, Molarity = 0.5M
Percentage weight of solute = 80%
density = d
Molar mass of solute = 40 g/mol
∴ 0.5 = 80 × 10 × d/40
⇒20 = 800 × d
⇒ 1/40 = d
∴ density of solution = 1/40 g/ml
Now, mass of solution = 1/40 × volume of solution
= 1/40 × 60000 g [ 1L = 1000ml ]
= 1500 g

Hence, amount of sample = 1500 g