Question -
Answer -
Molar mass of Na2CO3 =(2 × 23) + 12.00 + (3× 16)
= 106 g mol–1
Now, 1 mole of Na2CO3 means106 g of Na2CO3.
0.5mol of Na2CO3 
= 53 g Na2CO3
⇒ 0.50 M of Na2CO3 =0.50 mol/L Na2CO3
Hence, 0.50 mol of Na2CO3 ispresent in 1 L of water or 53 g of Na2CO3 is presentin 1 L of water.