Liquid steel contains initially 0.05 mass% P and this has to be reduced to 0.01 mass% using a basic slag. The equilibrium distribution ratio of P between slag and metal is $$Lp = \frac{{\left( {\% P} \right){\text{slag}}}}{{\left( {\% P} \right){\text{metal}}}}, = 80;$$ Assuming that in itially the slag does not contain any phosphorous (P), then the minimum weight of slag (ton) required per ton of steel is
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