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2 years ago

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6. A 0.1310 g sample of an unknown diprotic acid is diluted to 100.00 mL and titrated by using 0.1910 M NaOH. If 14.20 mL of the

NaOH solution is required to reach the second equivalence point, what is the molar mass of the acid?

1 answer:

miskamm [114]2 years ago

3 0

Answer:

$MM_{acid}=96.9g/mol$

Explanation:

Hello,

In this case, it is more convenient to define this titration in terms of normality:

$eq-g_{acid}=eq-g_{NaOH}\\N_{acid}V_{acid}=N_{NaOH}V_{NaOH}$

In such a way, the normality of the acid, considering the normality of sodium hydroxide equals its molarity (one hydroxile in its structure) is:

$N_{acid}=\frac{N_{NaOH}V_{NaOH}}{V_{acid}}=\frac{0.1910N*14.20mL}{100.00mL} \\N_{acid}=0.0271\frac{eq-g}{L}$

Thus, the moles are:

$n_{acid}=0.0271\frac{eq-g}{L}*0.10000L*\frac{1mol}{2eq-g} =1.356x10^{-3}mol$

Hence, the molar mass:

$MM_{acid}=\frac{0.1310g}{1.356x10^{-3}mol} \\MM_{acid}=96.9g/mol$

Best regards.

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Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

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Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

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Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

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Putting the values in 1

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x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

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2 years ago

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An amount of solid barium chloride, 20.8 g, is dissolved in 100 g water in a coffee-cup calorimeter by the reaction: BaCl2 (s) 

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Explanation :

First we have to calculate the heat gained by the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

m = mass of water = 100 g

c = specific heat = $4.04J/g^oC$

$T_{final}$ = final temperature = $26.6^oC$

$T_{initial}$ = initial temperature = $25.0^oC$

Now put all the given values in the above formula, we get:

$q=100g\times 4.04J/g^oC\times (26.6-25.0)^oC$

$q=646.4J$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles barium chloride = $\frac{\text{Mass of barium chloride}}{\text{Molar mass of barium chloride}}=\frac{20.8g}{208.23g/mol}=0.0998mole$

$\Delta H=-\frac{646.4J}{0.0998mole}=-6476.95J/mole=-6.48kJ/mole$

Therefore, the enthalpy change during the reaction is -6.48 kJ/mole

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2 years ago

Describe how to prepare the solution using a 250.0 mL volumetric flask by placing the steps in the correct order. Not all of the

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Answer:

The correct order will be

a. Transfer the measured amount of NaCl to the volumetric flask.

e. Dissolve the NaCl in less than 250 mL of water and mix well.

b. Dilute the solution with water to the 250.0 mL mark.

Explanation:

Preparation of NaCl solution in 250.0 ml volumetric flask:

Add the weighed NaCl directly to volumetric flask and add small amount of water to it and mix it will until all NaCl gets dissolved( if not add small water amount of water more)

After dissolving NaCl add the water upto the mark.

The correct order will be

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Answer:

C) 0.28 M

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

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Given :

<u>For Potassium hydroxide : </u>

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Volume = 40.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 40.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Potassium hydroxide is same as the moles of Potassium hydroxide as shown below:

$Moles =0.25 \times {40.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Potassium hydroxide = 0.01 moles

Barium hydroxide will furnish hydroxide ions as:

$Ba(OH)_2\rightarrow Ba^{2+}+2OH^-$

Given :

<u>For Barium hydroxide : </u>

Molarity = 0.15 M

Volume = 60.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 60.0×10⁻³ L

Thus, moles of hydroxide ions furnished by Barium hydroxide is twice the moles of Barium hydroxide as shown below:

$Moles =2\times 0.15 \times {60.0\times 10^{-3}}\ moles$

Moles of hydroxide ions by Barium hydroxide = 0.018 moles

Total moles = 0.01 moles + 0.018 moles = 0.028 moles

Total volume = 40.0×10⁻³ L + 60.0×10⁻³ L = 100×10⁻³ L

Concentration of hydroxide ions is:

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$Molarity_{OH^-}=\frac{0.028 }{100\times 10^{-3}}$

<u> The final concentration of hydroxide ion = 0.28 M</u>

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2 years ago

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