How Do You Calculate the Solubility of Sodium Sulfite?

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To calculate the mass percent concentration of sodium sulfite (Na2SO3) in a saturated solution at 20 degrees Celsius, the formula used is mass of Na2SO3 divided by the total mass of the solution. Given that the solubility is 27 g per 100 g of water, the total mass of the solution is 127 g, resulting in a mass percent of 21%. For the second question regarding the solubility of Na2SO3·7H2O in 50 g of water, the calculation involves accounting for the water in the hydrate. By using the mass percent from the first problem, it can be determined that 37 g of Na2SO3·7H2O is soluble in 50 g of water. Understanding these calculations is crucial for mastering solubility concepts in chemistry.
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I have a large exam tomorrow in chemistry and know how to do most problems on my practice test except these two

If the solubility of sodium sulfite (Na2SO3) is 27(g/100g water) at 20 degrees celcius, what is the mass percent concentration of the saturated solution of Na2SO3 at 20 degrees celcius?

How many grams of Na2SO3 * 7H2O is soluble in 50g water at 20 degrees celcius?



the answer to the first question is 21 and the 2nd answer is 37 but for the life of me I can't figure out how they got these answers.
 
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For the first one, mass percent of Na2SO3 = mass of Na2SO3 / total mass of solution. If you assume you have a saturated solution with 27 g Na2SO3, what's the total mass of the solution? Then you'll need this result for the next one.

The second one is a little tricky since you're working with the added water in Na2SO3·7H2O. Use the equation for finding the mass percent of Na2SO3 while using x for the moles of Na2SO3 and Na2SO3·7H2O; this mass percent of Na2SO3 must be equal to the mass percent you got for the first problem.
 
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