What is the value of x when 32.2 g of Na2SO4•xH2O are heated leaving 14.2 g of anhydrous Na2SO4? Mr(H2O) = 18; Mr(Na2SO4) = 142.Na2SO4•xH2O (s) → Na2SO4 (s) + xH2O (g)A. 0.1B. 1C. 5D. 10

The problem is asking for the value of x, which represents the number of water molecules in the hydrated sodium sulfate (Na2SO4•xH2O).

First, we need to find the mass of water that was lost when the hydrated sodium sulfate was heated. We can do this by subtracting the mass of the anhydrous sodium sulfate from the mass of the hydrated sodium sulfate:

32.2 g (hydrated sodium sulfate) - 14.2 g (anhydrous sodium sulfate) = 18 g (water)

Next, we need to find the number of moles of water that this mass represents. We can do this by dividing the mass of the water by the molar mass of water:

18 g (water) ÷ 18 g/mol (molar mass of water) = 1 mol (water)

Finally, we need to find the number of moles of sodium sulfate. We can do this by dividing the mass of the anhydrous sodium sulfate by the molar mass of sodium sulfate:

14.2 g (anhydrous sodium sulfate) ÷ 142 g/mol (molar mass of sodium sulfate) = 0.1 mol (sodium sulfate)

The value of x is the ratio of the number of moles of water to the number of moles of sodium sulfate:

x = 1 mol (water) ÷ 0.1 mol (sodium sulfate) = 10

So, the correct answer is D. 10.