Question -
Answer -
Molar mass of Na2CO3┬а=(2 ├Ч 23) + 12.00 + (3├Ч 16)
= 106 g molтАУ1
Now, 1 mole of Na2CO3┬аmeans106 g of Na2CO3.
0.5mol of Na2CO3┬а
= 53 g Na2CO3
тЗТ 0.50 M of Na2CO3┬а=0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3┬аispresent in 1 L of water or 53 g of Na2CO3┬аis presentin 1 L of water.