Solution.
m(BaCl2)=12.5g;m(BaCl_2)=12.5g;m(BaCl2)=12.5g;
V=250mL=0.25L;V=250mL=0.25L;V=250mL=0.25L;
w(BaCl2)=m(BaCl2)V;w(BaCl_2)=\dfrac{m(BaCl_2)}{V};w(BaCl2)=Vm(BaCl2);
w(BaCl2)=12.5g250mL=0.05g/mL;w(BaCl_2)=\dfrac{12.5g}{250mL}=0.05g/mL;w(BaCl2)=250mL12.5g=0.05g/mL;
Answer: w(BaCl2)=0.05g/mL.w(BaCl_2)=0.05g/mL.w(BaCl2)=0.05g/mL.
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