Answer:
(a) Benzene = 0.26; toluene = 0.74
(b) Benzene = 0.55
Explanation:
1. Calculate the composition of the solution
For convenience, let’s call benzene Component 1 and toluene Component 2.
According to Raoult’s Law,

where
p₁ and p₂ are the vapour pressures of the components above the solution
χ₁ and χ₂ are the mole fractions of the components
p₁° and p₂° are the vapour pressures of the pure components.
Note that
χ₁ + χ₂ = 1
So,

χ₁ = 0.26 and χ₂ = 0.74
2. Calculate the mole fraction of benzene in the vapour
In the liquid,
p₁ = χ₁p₁° = 0.26 × 75 mm = 20 mm
∴ In the vapour

Note that the vapour composition diagram below has toluene along the horizontal axis. The purple line is the vapour pressure curve for the vapour. Since χ₂ has dropped to 0.45, χ₁ has increased to 0.55.
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Answer:
- According to the law <br> Mass of reactants = mass of product, here <br> `underset(10 g)(CaCO_(3))rarr underset(4.4 g)(CO_(2))+underset(x)(CaO)` <br> Hence, x = 10 g - 4.4 g = 5.6 g <br> Which is mass of CaO.d
- In the first compound <br> Hydrogen = 5.93 % <br> Oxygen = `(100-5.93)% = 94.07 %` <br> In the second compound <br> Hydrogen = 11.2 % <br> Oxygen `= (100-11.2)%=88.8%` <br> In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts ...
- (The ratio of Cu combining with fixed weight of oxygen in black and red oxide is 1 : 2 respectively. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.) {Check something more in the above attachment!}
- Refer to the above attachment
Explanation:
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Answer:
0.0107 mol
Explanation:
Multiply concentration by volume (in liters) to get moles.
0.5 M • 0.0213 L = 0.0107 mol
Subtracting the mass of (flask+water) from the empty flask gives:
95.023 g - 85.135 g = 9.888 grams of water
Dividing this by the given volume of 10.00 mL water gives:
9.888 grams of water / 10.00 mL of water = 0.9888 g/mL of water
Therefore, based on this sample, the density of water is 0.9888 g/mL, which is close to the usually accepted approximation of 1 g/mL.