Answer:
Molar absorptivity or molar extinction co-effecient = 2120.14 cm⁻¹M⁻¹
Explanation:
First convert Concentration from ppm inM or mol/l
⇒ Molar mass of KMnO₄ = 158.03 g
⇒ 4.48 ppm = 4.48 mg/l = 4.48 x 10⁻³ g/l
⇒ Molarity = 
= 2.83 x 10⁻⁵ molar
Absorbance (A) = - log(T) ( T = % transmittance)
= - log(0.859)
= 0.06
According to Lambert Beer's law
ε = 
or, ε = 
or, ε = 2120.14 cm⁻¹M⁻¹
Where
ε = Molar absorptivity
A = absorbance
C = Molar concentration of KMnO₄ solution
l = length
True.
A mixture is composed of two or more pure substance that are physically combined but are not combined chemically and thus no electrons (of their atoms) are involved.
The primary properties of a mixture include:
1. The components of a mixture are easily separated
2. The components each keep their original properties.
3. The proportion of the components may vary.
There are two main categories of mixtures:
1. Heterogeneous mixtures - substances are not evenly distributed e.g. oil and water mixture
2. Homogeneous mixtures - substances are evenly distributed throughout the mixture e.g. salt water, air.
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
Answer:
The concentration of the dilute sample will be 0.361 g/ml
Explanation:
If a solution is diluted into 1:10 ratio then the amount of solute of that solution will be decreased by 10 times.
The initial concentration of the stock solution was 3.61g/ml but when the solution is diluted in 1:10 ratio the solute concentration is also decreased by 10 times.SO at present the solute concentration becomes 3.61/10=0.361 g/ml.
