If the mole fraction of k2so3 in an aqueous solution is 0.0328 what is the weight/weight % of k2so3
2 answers:
When we assume that our solvent is water with total weight 100 g: so we need to get the mass of K2SO3: mass of K2SO3 = moles K2SO3 * Mw of K2SO3 now we need to get first the Mw of K2SO3 = 2*39 + 1*32 + 3*16 = 158 g/mol by substitution: ∴ mass of K2SO3 = 0.0328 moles * 158 g/mol = 5.18 g ∴the mass of water = 100 g - 5.18 g = 94.8 g ∴ moles of water = mass of water / Mw of water = 94.8 g / 18 g/mol = 5.2 moles ∴the total moles = moles of water + moles of K2SO3 = 5.2 + 0.0328 = 5.266 moles∴ the weight/weight % of K2SO3 = moles of K2SO3 / total moles*100 = 0.0325 moles / 5.266 moles *100 = 0.0062 *100 = 0.62 %
Answer:
Explanation:
Hello,
In this case, since potassium sulfite is in an aqueous solution, that is water as solvent, one could assume 0.0328 moles of potassium sulfite and 1 mole of solution (total), therefore, the moles of water are:
Thus, one calculates each compound's grams by using their molar masses as:
Therefore, its weight/weight % turns out:
Best regards.
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