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arsen [322]
3 years ago
14

A solution of ethanol (c2h5oh) in water is prepared by dissolving 75.0 ml of ethanol (density 0.79 g/cm3 ) in enough water to ma

ke 250.0 ml of solution. What is the molarity of the ethanol in this solution?
Chemistry
1 answer:
navik [9.2K]3 years ago
5 0

Given the density of ethanol = 0.79\frac{g}{cm^{3} }

Volume of ethanol = 75.0 mL

Calculating the mass if ethanol from density and volume:

75.0 mL * \frac{0.79g}{1cm^{3} } *\frac{1cm^{3} }{1mL}=59.25g

Molar mass of ethanol = (2*atomic weight of C)+(6*atomic weight of H)+(1*atomic weight of O)

                                     =(2*12g.mol) + (6*1g/mol) + (1*16g/mol)

                                     =46g/mol

Moles of ethanol = 59.25 g *\frac{1 mol}{46g} =1.288 mol

Volume of the solution = 250.0 mL

Converting the volume from mL ot L:

250.0mL*\frac{1L}{1000mL}=0.2500L

Molarity of ethanol in the solution = \frac{1.288mol}{0.250L} =5.15 M

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3 years ago
Does anyone know how to do this?
Sloan [31]

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find out the number of moles and use the molar ratio (numbers in front of formulas (in this case they are all 1) to determine how many moles of each product you are going to get theoretically

n=m/M is the equation to use to get moles here

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4 0
3 years ago
A 0.25-mol sample of a weak acid with an unknown Pka was combined with 10.0-mL of 3.00 M KOH, and the resulting solution was dil
Masteriza [31]

Answer : The value of pK_a of the weak acid is, 4.72

Explanation :

First we have to calculate the moles of KOH.

\text{Moles of }KOH=\text{Concentration of }KOH\times \text{Volume of solution}

\text{Moles of }KOH=3.00M\times 10.0mL=30mmol=0.03mol

Now we have to calculate the value of pK_a of the weak acid.

The equilibrium chemical reaction is:

                          HA+KOH\rightleftharpoons HK+H_2O

Initial moles     0.25     0.03        0

At eqm.    (0.25-0.03)   0.03      0.03

                     = 0.22

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[HK]}{[HA]}

Now put all the given values in this expression, we get:

3.85=pK_a+\log (\frac{0.03}{0.22})

pK_a=4.72

Therefore, the value of pK_a of the weak acid is, 4.72

7 0
3 years ago
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