Question

Calculate the amount of sodium carbonate required to make 100cm3 of a 0.25M solution.

Answer 1

Answer:

The answer is 2.65 gm

Explanation:

According to the question, amount of Sodium Carbonate or Na2CO3 is needed to prepare 0.25 Molar solution when the given volume is 100 cm3 which can be written as 100mL.

So, we have

Volume = 100mL= 0.1 L

Molarity or concentration = 0.25M

Now if we apply the formula

$mole \: number \: = molarity \: \times volume \: (in \: liter)$

we get mole number = 0.25 x 0.1 = 0.025 mole

Now,

For 1 mole Na2CO3 = 106gm

Thus 0.025 mole Na2CO3 = (106 x 0.025) = 2.65 gm

I hope it helped.