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3 years ago

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The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc

ulate the percentage by which the apparent concentration deviates from the known concentration.

1 answer:

Korvikt [17]3 years ago

6 0

Answer:

The percentage deviation is $\Delta M = 1.87$ %

Explanation:

From the question we are told that

The concentration is of the solution is $C = 1.0*10^{-5} M$

The true absorbance A = 0.7526

The percentage of transmittance due to stray light $z = 0.56$ % $=\frac{0.56}{100} = 0.0056$

Generally Absorbance is mathematically represented as

$A = -log T$

Where T is the percentage of true transmittance

Substituting value

$0.7526 = - log T$

$T = 10^{-0.7526}$

$= 0.177$

$= 17.7$ %

The Apparent absorbance is mathematically represented

$A_p = -log (T +z)$

Substituting values

$A_p = -log(0.177 + 0.0056)$

$= -log(0.1826)$

= 0.7385

The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

$\Delta A = \frac{A -A_p}{A} * \frac{100}{1}$

$= \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}$

$\Delta A = 1.87$ %

Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration is

$\Delta M = 1.87$ %

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<u>For a:</u>

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$1.)C_2H_5OH(l)(35^oC)\rightarrow C_2H_5OH(l)(78^oC)\\2.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

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We are given:

Change in temperature remains the same.

$m=42.0g\\c_l=2.3J/g.K\\T_2=78^oC\\T_1=35^oC\\\Delta T=[T_2-T_1]=[78-35]^oC=43^oC=43K$

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Conversion factor: 1 kJ = 1000 J

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Hence, the total heat required is 36621.5 J

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The processes involved in the given problem are:

$1.)C_2H_5OH(s)(-155^oC)\rightarrow C_2H_5OH(s)(-144^oC)\\2.)C_2H_5OH(s)(-144^oC)\rightarrow C_2H_5OH(l)(-144^oC)\\3.)C_2H_5OH(l)(-144^oC)\rightarrow C_2H_5OH(l)(78^oC)\\4.)C_2H_5OH(l)(78^oC)\rightarrow C_2H_5OH(g)(78^oC)$

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We are given:

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We are given:

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