Correct Answer - 0.125 M
Setp I. Calculation of mass of `CaCI_(2)` present in solution.
lonisation of `CaCI_(2)` and no. of different species in the soluion may be represented as follows : `CaCI(s)overset((aq))toca^(2+)(aq)+2CI^(-)(aq)`
`(1.505xx10^(22)"molecules")(1.505xx10^(22))(3.01xx10^(22))`
`6.022xx10^(22)"molecules of"CaCI_(2) " correspond to mass"=(1.505xx10^(22))/(6.022xx10^(23))xx(111g)=2.774 g`
Step II. Calculation of molarity of solution. `"Molarity of soluion (M)"=("Mass of "CaCI_(2)//"Molar mass")/("Volume of solution in Litress")`
`((2.774g)//(111" g mol"^(-1)))/(200//1000L)=0.125" mol L"^(-1)=0.125 M`