1.

200.0 mL of calcium chloride solution sotains 3.10xx10^(22)CI^(-) inoc. Calculate the molarity of the solution. Assume that calcium cholride is complately ionised.

Answer»


Solution :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`


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