What volume, in L, of a 6.3 M K2SO4 solution contains 44.1 g of K2SO4?
1 answer:
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Explanation:
Normal moles of = volume × normal concentration
= 4.7 × 0.139 = 0.6533 mol
Moles of in hyponatremia blood = volume × hyponatremia concentration
= 4.7 × 0.116 = 0.5452 mol
Moles of NaCl to be added = moles of extra needed
= 0.6533 mol - 0.5452 mol = 0.1081 mol
Mass of NaCl = moles × molar mass of NaCl
= 0.1081 mol × 58.443
= 6.317g
= 6.32 g (approx)
Thus, we can conclude that mass of sodium chloride would need to be added to the blood is 6.32 g.