2 mL of heptane plus 2 mL of alpha pinene give 4 mL ofsolution. How many grams p

Question

2 mL of heptane plus 2 mL of alpha pinene give 4 mL ofsolution. How many grams per mL are in that mixture? Suppose youdiluted 0.25 mL of the 50:50 heptane/alpha-pinene mixture to 5 mLwith solvent and measured the optical rotation. How many grams permL in the solution? By what factor would you have to multiply thevalue D to get the specific rotation ofalpha-pinene? 2 mL of heptane plus 2 mL of alpha pinene give 4 mL ofsolution. How many grams per mL are in that mixture? Suppose youdiluted 0.25 mL of the 50:50 heptane/alpha-pinene mixture to 5 mLwith solvent and measured the optical rotation. How many grams permL in the solution? By what factor would you have to multiply thevalue D to get the specific rotation ofalpha-pinene?

Explanation / Answer

-pinene :
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Density of -pinene d = 0.858 g/mL.
Volume of the - pinene in the mixture , v = 2 mL
So mass of - pinene , m = d * v = 1.716 g
1.716 g of - pinene is present in 4 mL of the solutionmixture
Therefore the the mass of -pinene present in 1 mL of thesolution is m' =1.716 / 4
                                                                                                           = 0.429 g
after dilution ,
From dilution law MV = M'V' M &M' -- Concentration mass of thesolute
V & V' ----- Volume
0.429g * 0.25 mL = m"" * 5mL       Since we are diluting thesolution to 5 mL
                        m""= 0.02145 g of - pinene
heptane :
----------
Density of heptane d' = 0.684 g/mL.
Volume of the heptane in the mixture ,v' = 2 mL
So mass of heptane, m'' = d' * v'= 1.368 g
1.368 g of heptane is present in 4 mL of the solution mixture
Therefore the the mass of heptane present in 1 mL of thesolution is m'" =1.368 / 4
                                                                                                           = 0.342 g From dilution law M'''V'' = M''''V''' 0.342g * 0.25 mL = m"" * 5mL       Since we are diluting thesolution to 5 mL
                        m""= 0.0171 g of heptane 0.342g * 0.25 mL = m"" * 5mL       Since we are diluting thesolution to 5 mL
                        m""= 0.0171 g of heptane

We know specific rotation s = /lC where = optical rotation           l= path length = length of the cell = 1cm          C= concentration of - pinene in g/100 mL = 0.02145g in 5mL                                                                              = (0.02145 * 100 ) / 5 g in 100 mL                                                                                = 0.429 g / 100mL Therefore s = (1/lC )                  = 2.33

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