Question

The amount of anhydrous na2co3 present in 250ml of 0.25m solution is

Answer 1

0.25M solution contains 0.25 moles in 1 litre (1000 ml)

250 ml of the solution will contain = 250/1000 x 0.25 = 0.0625 moles

Molar mass of Na2CO3

= 23x2 + 12 + 16x3

= 106 g/mol

Mass = Number of moles x molar mass

Mass of Na2CO3 = 0.0625 x 106

                            = 6.625 g

Answer 2

Answer: 6.625 grams of anhydrous $Na_CO_3$ is present in the solution.

Explanation: We are given a solution having molarity 0.25M

Given:

Molarity = 0.25M

Volume of the solution = 250 mL

To calculate the amount of solute, we use the formula:

$Molarity=\frac{w\times 1000}{M\times V(\text{ in mL})}$

Where, w = amount of solute present in the solution.

M = molar mass of the solute

V = volume of the solution

Finding the molar mass of $Na_2CO_3$

M = (2 × 23) + 12 + (3 × 16) = 106 g/mol

Putting the values, in the molarity equation, we get

$0.25=\frac{w\times 1000}{106\times 250}$

w = 6.625 grams

Amount of $Na_2CO_3$ present in the solution is 6.625 grams.