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How will you prepare 0.25 m `CaVI_(2)` solution? |
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Answer» Correct Answer - 27.75 g `"Molality of solution"=("No. of moles of"CaCI_(2))/("Mass of solvent in kg ")` `"No. of moles of "CaCI_(2)="Molality"xx"Mass of solvent in kg"` `=(0.25" mol lg"^(-1))xx(1 kg )= 0.25 mol` `"Molar mass of "CaCI_(2)=40+2xx35.5=111 " g mol"^(-1)` `"Mass of "CaCI_(2)=(0.25 mol)xx(111 g mol^(-1))=27.75 g.` |
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