1000 kg liquid steel at 0.03 wt.% S is desulphurized to 0.015 wt.% S. Assume equilibrium and initially sulfur-free slag. The slag/steel sulfur partition ratio, (wt.% S in slag)/[wt.% S in steel], is 50. Find slag required in kg, nearest integer.
The removed sulfur mass is $1000(0.03-0.015)/100=0.15$ kg. At the target steel concentration, slag contains $50(0.015)=0.75$ wt.% S, or mass fraction 0.0075. Slag mass is $0.15/0.0075=20$ kg, neglecting the very small change in bulk steel mass as in the intended balance.
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