$(a)\;4.2g\qquad(b)\;2.1g\qquad(c)\;3.81g\qquad(d)\;4.38g$

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Meq of $CaCO_3$=Meq of HCL=Meq of $CaCl_2$ formed=$525\times \large\frac{1}{10}$$=52.5$

$CaCl_2$ converted to $CaSO_4$ and then to plaster of paris($CaSO_4.\large\frac{1}{2}$$H_2O)$

$\therefore$ Meq of $CaCl_2$ formed=Meq of $CaSO_4$ obtained by $CaCl_2$

$\therefore$ Meq of $CaSO_4.\large\frac{1}{2}$$H_2O$ =Meq of $CaSO_4$ obtained by $CaCl_2$

$\large\frac{w}{\Large\frac{145}{2}}$$\times 1000=52.5$

Weight of $CaSO_4\large\frac{1}{2}$$H_2O=3.81g$

Hence (c) is the correct answer.

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