1.

A solution of HCI is prepared by dissolving 5.5 g HCI in 200 g ehanol. The density of the solution is 0.79 g `mL^(-1)` Molarity of solution is :A. 0.58 MB. 0.16 MC. 0.92 MD. 1.2 M

Answer» Correct Answer - a
`"Molarity (M)"=("Mass of HCI//Molar mass")/("Volame of solution in litrese")`
Mass of HCI solution = 200+5.5=205.5 g
Density of solution = 0.79 g `mL^(-1)`
`"Volume of solution"=((205.5g))/((0.79gmL^(-1)))`
=260L
`"Molarity"=((5.5g)//(36.5g mol^(-1)))/(0.260L)`
=0.58 `mL^(-1)=0.58 M`


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