Question

What volume (in ML) of 12.0M HCL is needed to contain 3.00 moles of HCL?

Answer 1

\[c_i=\frac{n_i}{V}\]\[V=\frac{n_i}{c_i}\]\[V=\frac{3.00mol(HCl)}{12.0mol*L^{-1}(HCl)}=0.250L (HCl)\]\[0.250L(HCl)*\frac{1000mL}{1L}=250.mL(HCl)\]

Answer 2

thanks