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crimeas [40]
3 years ago
10

A student prepares a weak acid solution by dissolving 0.2000 g HZ to give 100. mL solution. The titration requires 33.5 mL of 0.

1025 M NaOH. Calculate the molar mass of the acid. WebAssign will check your answer for the correct number of significant figures. g/mol (a) Would the molar mass be too high, too low, or unaffected if the student accidentally used 0.1000 g in the calculation
Chemistry
1 answer:
nexus9112 [7]3 years ago
3 0

Answer:

a)M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88 g/mol

b) if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

Explanation:

0.2000 gHZ gives  100ml acid solution

33.5 ml of 0.1025 M NaOH  is required to prepare it

the moles = mass / molar mass

mass = 0.200 gHZ

moles = 0.0335*100 * 0.1025 = 0.034

therefore molar mass = mass / moles

                                  M=0.20/(0.335*0.1025)= 0.20/ 0.034 = 5.88

if 0.100g is used instead of 0.200g

M = 0.1 / 0.034 = 2.94  hence the molar mass will be too low

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Two aqueous sulfuric acid solutions containing 20.0 wt% H2SO4
kozerog [31]

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The feed ratio (liters 20%  solution/liter 60% solution) is 3,08

Explanation:

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20 kg H₂SO₄ × (1 kmol/98,08 kg) = 0,2039 kmol H₂SO₄≡ <em>203,9 mol</em>

The final molarity 4,00M comes from:

\frac{203,9 moles+ Xmoles}{87,8L + Yliters} <em>(1)</em>

Where X moles and Y liters comes from 60,0 wt% H₂SO₄

100 L of 60,0 wt% H₂SO₄ are:

100L×\frac{1,498 kgsolution}{L}×\frac{60 kg H_{2}SO_{4}}{100kgSolution}×\frac{1kmol}{98,08kg H_{2}SO_{4}} = <em>0,9164 kmolH₂SO₄ ≡ 916,4 moles</em>

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Replacing (2) in (1):

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Thus, feed ratio (liters 20%  solution/liter 60% solution):

87,8L/28,52L = <em>3,08</em>

I hope it helps!

8 0
3 years ago
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