NEET-XII-Chemistry
c2022 year:2022
- #26What mass of `` 95 \% `` pure `` CaCO _{3} `` will be required to neutralise `` 50 \, mL `` of `` 0.5\, M HCl `` solution according to the following reaction? `` CaCO _{3( s )}+2 HCl _{( aq )} \rightarrow CaCl _{2( aq )}+ CO _{2( g )}+2 H _{2} O _{( l )} `` [Calculate upto second place of decimal point]
(A) `` 9.50\, g ``
(B) `` 1.25 \, g ``
(C) `` 1.32\, g ``
(D) `` 3.65 \, g ``digAnsr: CAns : `` CaCO _{3( s )}+2 HCl _{\text {(aq.) }} \rightarrow CaCl _{2 \text { (aq.) }}+ CO _{2( a )}+ H _{2} O _{(0)} ``
no. of moles of `` CaCO _{3} `` (pure) `` =\frac{1}{2} \times `` mole of `` HCl ``
`` \text { [Mole }=\text { molarity } \times \text { volume(in ltr. })] ``
`` =\frac{1}{2} \times 0.5 \times \frac{50}{1000}=0.0125 ``
weight of `` CaCO _{3} `` (pure) `` = `` mole `` \times `` mol. wt
`` =0.0125 \times 100=1.25\,g ``
`` \% `` purity `` =\frac{\text { wt. of pure substance }}{\text { wt. of impure sample }} \times 100 ``
`` 95=\frac{1.25}{\text { wt. of impure sample }} \times 100 ``
wt. of impure sample `` =\frac{1.25 \times 100}{95}=1.32\, g ``