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# Mole fraction problem?

If the mole fraction of K2SO3 in an aqueous solution is .0328, what is the weight/weight % (percent by mass) of K2SO3?

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This concentration unit focuses on the solute as a percent (by mass) of the total solution.  Since to weight of the total solution is not given, I will assume 100g, as an example.

One mole of K2SO3 will have potassium (2x39), sulfur (1x32), and oxygen (3x16) = 158g/mole

So 158g/mole x 0.0328mole = 5.2g

100g of solution - 5.2g os solute = 94.8g  of water รท 18g/mple = 5.2 moles of water
the total moles present = 5.2 + 0.0328= 5.266

The w/w % will be 0.0325/5.266 = 0.623%

If the total weight of the solution is different than 100g, the percentage will be different.