In: Chemistry
A chemist prepared a solution by dissolving 52.0 g of hydrated sodium carbonate in water to a total volume of 5.00 dm3. The concentration was determined to be 0.0366 M. Determine the formula of the hydrated sodium carbonate.
Moleculer weight = (wt/molarity)*(1/Vol in lt)
= (52/0.0366)*(1/1)
= 1420.76
Unhydrated Na2CO3 mol.wt is = 106 gm/mol
Subtract Unhydrated moleculer weight from hydrated moleculer weight
= 1420.76 - 106 = 1314.76
Divide the above value with moleculer weight of water
= 1314.76/18.015 = 72.98
so no.of H2O molecules are 73
the formula is Na2CO3 73*H2O