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Calculate the molarity of each of the following solutions: 30g of `Co(NO_(3))_(2)6H_(2)O` in 4.3L of solution `30 mL of0.5 M H_(2)SO_(4)` diluted to 500 mL |
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Answer» `"Molarity of solution"=("Mass of solute/Molar mass of solute")/("Volume of solution in litres")` Molar mass solute, `Co(NO_(3))_(2).6H_(2)O=30 g`. Molar mass of solute, `Co(NO_(3))_(2).6H_(2)O=59+2xx14+6xx16+6xx18=219 g mol^(-1)` Volume of solution=4.3 L `"Molarity(M)"=((30g)//(291g mol^(-1)))/((4.3L))=0.024 molL^(-1)=0.024 M` Volume of undiluted `H_(2)SO_(4)` solution `(V_(1))`=30 mL Molarity of undiluted `H_(2)SO_(4)` solution `(M_(1))`=1.5 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)` `M_(2)=(M_(1)V_(1))/V_(2)=((30mL)xx(0.5M))/((500mL))=0.03 M` |
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