Answer:
A is denser than B as it's volume for the same mass is smaller.
Explanation:
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In this case, we first need to take into account that the density of each metal A and B is computed by dividing the mass over the volume of each metal which is actually computed by substracting the volume of water from the volume of the water and the solid:
Next, we compute the densities as shown below:
In such a way, A is denser is B as it's volume for the same mass is smaller.
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Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.
Explanation : Given,
Vapor presume of 1‑Propanol = 20.9 torr
Vapor presume of 2‑Propanol = 45.2 torr
Mole fraction of 1‑Propanol = 0.540
Mole fraction of 2‑Propanol = 1-0.540 = 0.46
First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.
where,
= partial vapor pressure of 1‑Propanol
= vapor pressure of pure substance 1‑Propanol
= mole fraction of 1‑Propanol
and,
where,
= partial vapor pressure of 2‑Propanol
= vapor pressure of pure substance 2‑Propanol
= mole fraction of 2‑Propanol
Thus, total pressure = 11.3 + 20.8 = 32.1 torr
Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.
and,
Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.