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Posted by on Monday, April 7, 2008 at 4:53pm.

A solution is made by mixing 25.0 mL of toluene C6H5CH3 d=0867gmL with 135.0 mL of benzene C6H6 d=0874gmL . Assuming that the volumes add upon mixing, the molarity (M) and molality (m) of the toluene are

how would i start this problem?

  • Chemistry - , Monday, April 7, 2008 at 9:24pm

    Start by convert 20.0 mL toluene to mass, then convert that to mols. mols = g/molar mass.
    Then convert benzene to mass.
    molality toluene = mols toluene/kg benzene.

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