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Nana76 [90]
3 years ago
15

Initially a NaOH solution was standardized by titration with a sample of potassium hydrogenphthalate, KHC8H4O4, a monoprotic aci

d often used as a primary standard. A sample of pureKHC8H4O4 weighing 1.518 grams was dissolved in water and titrated with the NaOH solution. Toreach the equivalence point, 26.90 milliliters of base was required. Calculate the molarity of theNaOH solution. (Molecular weight: KHC8H4O4 = 204.2)
Chemistry
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

Molarity of the NaOH solution is 0.2764 M.

Explanation:

Mass of potassium hydrogenphthalate = 1.518 g

Moles of potassium hydrogenphthalate =\frac{1.518 g}{204.2 g/mol}=0.007434 mol

KHC_8H_4O_4+NaOH\rightarrow NaKC_8H_4O_4+H_2O

According to reaction, 1 mole of potassium hydrogenphthalate reacts with 1 mole of NaOH, then 0.007434 moles pf potassium hydrogenphthalate will :

\frac{1}{1}\times 0.007434 mol=0.007434 of NaOH

Moles of NaOH = 0.007434 mol

Volume of NaOH solution =  26.90 mL = 0.02690 L

1 mL = 0.001 L

Molarity=\frac{Moles}{Volume (L)}

Molarity of the NaOH solution :

=\frac{0.007434 mol}{0.02690 L}=0.2764 M

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Answer:

The molecular weight is Z =  111.2 \ g/mol

Explanation:

From the question we are told that

   The mass of the sample is  m =  0.98 \  g

    The temperature is  T  =  348 K

    The volume which the gas occupied is  V  =  265 \ ml  = 265 *10^{-3} L

     The pressure is  P  =  0.95 \  atm

Generally from the ideal gas equation we have that

       PV  =  n RT

Here n is the number of moles of the gas while the R is the gas constant with value  R  =  0.0821 \ atm \cdot L  \cdot mol^{-1} \cdot K^{-1}

        n = \frac{PV}{ RT}

=>      n = \frac{ 0.95 * 265 *10^{-3} }{   0.0821 * 348}

=>      n = 0.00881 \  mol

Generally the molecular weight is mathematically represented as

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Student A performed gravimetric analysis for sulfate in her unknown using the same procedures we did . Results of her three tria
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(ii) % mistake = (67.6 - 66.2)/67.6 = 2.07  

(iii) % mistake = (67.6 - 67.1)/67.6 = 0.74  

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(iii) % mistake = (67.6 - 66.5)/67.6 = 1.63  

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Answer:

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<em>The balanced equation for the chemical reaction between solid calcium and iron (III) chloride is given below as:  </em>

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