`6.0g` sample of `CaCO_(3)` reacts with 20g solution of HCl having 20% by mass of HCl (density `=1.10g"/"mL`). Calculate percentage purity of `CaCO_(3)` sample .

Options

A. `9.13%`
B. `91.33%`
C. `54.8%`
D. `5.48%`

Answer

B

Explanation

`V_("HCl solution")= (20)/(1.1)= 18.18cm^(2)" "N_("HCl")= (W_("HCl")xx 1000)/(M_("HCl")xx V_("HCl"))=(4xx 1000)/(36.5xx 18.18)`
(Since 20% of 20g = 4g) `" "` Meq. Of HCl = Meq.of `CaCO_(3)`
`(4xx1000)/(36.5xx 18.18)xx 18.18= 109.6" "W_(CaCO_3)= ("Meq. of"CaCO_(3))/(1000)xx(E_(CaCO_3))/(2)`
`=(109.6)/(1000)xx(100)/(2)= 5.48g" "` % purity of sample `=(5.48)/(6)xx100= 91.33%`.