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dlinn [17]
2 years ago
5

What is the concentration (m) of ch3oh in a solution prepared by dissolving 16.8 g of ch3oh in sufficient water to give exactly

230 ml of solution?
Chemistry
1 answer:
Natalija [7]2 years ago
5 0

The Molarity concentration is expressed in units of moles / L. So let us first determine the number of moles of CH3OH, and then divide that amount by the total volume of 0.230 L of solution.<span>

<span>To determine the number of moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight of CH3OH: (MW of CH3OH = 32 g / mol)</span>

number of moles = 16.8 g / (32 g / mol)</span>

number of moles = 0.525 mol CH3OH <span>

Then we calculate for molarity:</span>

Molarity = 0.525 mol CH3OH / .230 L

Molarity = 2.2826 mol / L

<span>Molarity = 2.28 M</span>

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Answer:

A i. Internal energy ΔU = -4.3 J ii. Internal energy ΔU = -6.0 J B. The second system is lower in energy.

Explanation:

A. We know that the internal energy,ΔU = q + w where q = quantity of heat and w = work done on system.

1. In the above q = -7.9 J (the negative indicating heat loss by the system). w = 3.6 J (It is positive because work is done on the system). So, the internal energy for this system is ΔU₁ = q + w = -7.9J + 3.6J = -4.3 J

ii. From the question q = +1.5 J (the positive indicating heat into the system). w = -7.5 J (It is negative because work is done by the system). So, the internal energy for this system is ΔU₂ = q + w = +1.5J + (-7.5J) = +1.5J - 7.5J = - 6.0J

B. We know that ΔU = U₂ - U₁ where U₁ and U₂ are the initial and final internal energies of the system. Since for the systems above, the initial internal energies U₁ are the same, then we say U₁ = U. Let U₁ and U₂ now represent the final energies of both systems in A i and A ii above. So, we write ΔU₁ = U₁ - U and ΔU₂ = U₂ - U where ΔU₁ and ΔU₂ are the internal energy changes in A i and A ii respectively. Now from ΔU₁ = U₁ - U, U₁ = ΔU₁ + U and U₂ = ΔU₂ + U. Subtracting both equations U₁ - U₂ = ΔU₁ - ΔU₂

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