~Solution :-
Here, we have been given that;
- $ \bf{1 ml}$ = 1.17 g
- $ \bf{1 l}$ = 1170 g
Hence, we can note that,
- No. of moles of HCL (n) = $ \sf{\frac{1170}{36.5}}$
(n) = 32.05
[tex] \\ [/tex]
Thus,
Hence, molarity of the given solution will be 32.05 mg/L.
