Solution.
"m(BaCl_2)=12.5g;"
"V=250mL=0.25L;"
"w(BaCl_2)=\\dfrac{m(BaCl_2)}{V};"
"w(BaCl_2)=\\dfrac{12.5g}{250mL}=0.05g\/mL;"
Answer: "w(BaCl_2)=0.05g\/mL."
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