How are `0.50 mol Na_(2)CO_(3)` and `0.50 M Na_(2)CO_(3)` different?

Explanation

Molar mass of `Na_(2)CO_(3)=(2xx23)+12.00+(3xx16)`
= 106 g` mol^(-1)`
Now , 1 mole of `Na_(2)CO_(3)` Means 106 g of `Na_(2)CO_(3)` .
`therefore` 0.5 mol of `Na_(2) CO_(3) =(106 g)/(1 "mole")xx0.5 mol Na_(2)CO_(3)`
= 53 g `Na_(2)CO_(3)`
`implies 0.50 M of Na_(2)Co_(3)=0.50|mol//L Na_(2)CO_(3)`
Hence , 0.50 mol of `Na_(2)CO_(3)` is present in 1 L of water or 53 g `Na_(2)CO_(3)` is present in 1 L of water .