We will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
I believe that the answer is D.
I hope this helps. :)
In terms of a deeper scientific reason, I am not sure, but the basic reason is quite simple. "Mud" tends to look like a mix between a solid, dirt, and a liquid, water or some other liquid. Since it is, in fact, a cross between a solid and a liquid, it has properties of both. It has certain physical and visual properties that only a solid would have, such as texture and opaqueness, but it also has physical properties of a liquid. Since it leans more towards the liquid side than the solid side, we say mud "flows" rather than saying that it "rolls" or "bounces".
Answer:
[S₂] = 1.27×10⁻⁷ M
Explanation:
2 H₂S(g) ⇄ 2 H₂(g) + S₂(g), Kc=1,625x10⁻⁷
The equation of this reaction is:
1,625x10⁻⁷ = ![\frac{[H_2]^2[S_2]}{[H_{2}S]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_%7B2%7DS%5D%5E2%7D)
The equilibrium concentrations are:
[H₂S] = 0,162 - 2x
[H₂] = 0,184 + 2x
[S₂] = x
Replacing:
1,625x10⁻⁷ = ![\frac{[0,184+2x]^2[x]}{[0,162-2x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0%2C184%2B2x%5D%5E2%5Bx%5D%7D%7B%5B0%2C162-2x%5D%5E2%7D)
Solving:
4x³ + 0,736x² + 0,033856x - 4,3x10⁻⁹
x = 1.27×10⁻⁷
Thus, concentration of S₂ is:
<em>[S₂] = 1.27×10⁻⁷ M</em>
