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Effectus [21]
3 years ago
7

A bottle of wine contains 9.81 grams of C2H5OH, dissolved in 87.5 grams of water. The final volume of the solution is 100.0 mL.

Calculate the concentration of ethanol in wine in terms of molarity and mass percent.
Chemistry
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)

Explanation:

1. Molarity = moles solute / Volume solution in Liters

=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH

=> volume of solution (assuming density of final solution is 1.0g/ml) ...

volume solution =  9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution

Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH

2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)

From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln

= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.

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a sample of an oxide of antimony (sb) contain 19.75 g of antimony combine with 6.5 g of oxygen . what is the simplest formula fo
frez [133]

Explanation:

The given data is as follows.

      Mass of antimony = 19.75 g

      Molar mass of Sb = 121.76 g/mol

Therefore, calculate number of moles of Sb as follows.

                    Moles of Sb = \frac{mass}{\text{molar mass}}

                                         = \frac{19.75 g}{121.76 g/mol}

                                         = 0.162 mol

Mass of oxygen given is 6.5 g and molar mass of oxygen is 16 g/mol. Hence, moles of oxygen will be calculated as follows.

           Moles of oxygen = \frac{mass}{\text{molar mass}}

                                         = \frac{6.5 g}{16 g/mol}

                                         = 0.406 mol

Hence, ratio of moles of Sb and O will be as follows

                          Sb : O

                      \frac{0.162}{0.162} : \frac{0.406}{0.162}

                           1 : 2.5

We multiply both the ratio by 2 in order to get a whole number. Therefore, the ratio will be 2 : 5.

Thus, we can conclude that the empirical formula of the given oxide is Sb_{2}O_{5}.

4 0
3 years ago
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