Question

prepare the standard solution of 250ml 0.5 M Na2CO3

Answer 1

We have to prepare the standard solution of 250ml , 0.5M of $Na_2CO_3$ :

molar weight of $Na_2CO_3$= 2 × 23 + 12 + 3 × 16 = 46 + 60 = 106g
mole = molarity × volume/1000
= 0.5 × 250/1000 = 0.5/4 = 0.125
mass of $Na_2CO_3$ = 0.125 × 106g
= 13.25g

Use an electronic balance, we will weigh about 13.25g of $Na_2CO_3$ crystals in a weighing bottle and transfer these into a 250ml beaker.

wash the bottle two - three times, using distilled water and transfer all the washings into the beaker. Dissolve $Na_2CO_3$ crystals in the beaker by gentle stirring with a clean glass rod.

when $Na_2CO_3$ crystals dissolve completely in the beaker, transfer entire solution into 250ml standard flask through a funnel , using a glass rod.

wash the beaker two to three times with distilled water and transfer all the washings into the standard flask and finally wash the funnel thoroughly with distilled water.  This transfers the solution completely into the standard flask.

now use a wash bottle, add enough distilled water to the standard flask so that the level is just below the calibration mark on it. Add the last few drops of distilled water with a pipette until the lower level of the meniscus just touches the mark on the standard flask.

stopper the measuring flask and shake gently to make the solution uniform throughout.

Answer 2

Prepare the standard solution of 250ml 0.5M Na2CO3.

Answer:

Let's solve the answer for this.

We need the molarity of Sodium Carbonate.

so the formula morality is:

number of moles of the solute/volume of the the solution in litres.

Molarity= 0.5 M

Volume of the solution is = 250 ml

The molecular mass of sodium carbonate is = 23*2+12+16*3= 106

0.5= x*1000/ 106*250

X= 0.5*106*250/1000

X= 13.25 grams.

Answer: 13.25 grams.