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Calculate the molarity of each of the following solutions `:` `a. 30g` of `Co(NO_(3))_(2).6H_(2)O` in `4.3L` of solution `b. 30mL` of `0.5 M H_(2)SO_(4)` diluted to `500mL`. |
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Answer» (a) Molarity of solution `= ("Mass of solute/Molar mass of solute")/("Volume of solution in litres")` Mass of solutem `Co(NO_(3))_(2).6H_(2)O=30g` Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=291 "g mol"^(-1)` Volume of solution `=4.3 L` Molarity `(M)=((30g)//(291 "g mol"^(-1)))/((4.3L))=0.024"mol L"^(-1)=0.024M` (b) Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))=30` mL Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))=0.5` M Volume of diluted `H_(2)SO_(4)` solution `(V_(2))=500` mL Molarity of diluted `H_(2)SO_(4) (M_(2))` can be calculated as : `M_(1)V_(1)=M_(2)V_(2)` or `M_(2)=(M_(1)V_(1))/(V_(2))=((30mL)xx(0.5M))/((500mL))=0.03M`. |
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