If the mole fraction of k2so3 in an aqueous solution is 0.0328 what is the weight/weight % of k2so3

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 %