Respuesta :

~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

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Thus,

  • Molarity = 32.05 mg/L

Hence, molarity of the given solution will be 32.05 mg/L.

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