A chemist prepared a solution by dissolving 52.0 g of hydrated sodium carbonate in water to...

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.

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Expert Solution

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

queen_honey_blossom answered 1 month ago

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