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If the density of methanol is `0.793 kg L^(-1)` what ia its volume needed for making 2.5 L of its `0.25 M` solution? |
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Answer» Molar mass of methanol `(CH_(3)OH)` `=32 g mol^(-1)=0.032 kg mol^(-1)` Molarity of the given solution `=(W_(2) "in" kg)/(Mw_(2)xxV_(sol)(L))=(d_(sol)(kgL^(-1)))/(Mw_(2)(kg))` `=(0.793 kg L^(-1))/(0.032 kg mol^(-1))=24.78 M` Applying `underset(("Given solution"))(M_(1)xxV_(1))=underset("prepared)")underset("(Solution to be")(M_(2)V_(2)` `24.78xxV_(1)=0.25xx2.5 L` or `V_(1)=0.02522 L=25.22 mL` |
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