Question

A Knudsen cell containing a crystal (M = 157 g mol$^{-1}$) is carefully weighed and placed in an evacuated chamber thermostatted at 90 $^\circ$C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90 $^\circ$C in Pa from the fact that the weight loss is 39.2 mg.

Name: A Knudsen cell containing a crystal (M = 157 g mol^-1) is carefully weighed and placed in an evacuated chamber thermostatted at 90 ^∘C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90 ^∘C in Pa from the fact that the weight loss is 39.2 mg.
Uploaded: 2023-06-06T11:24:52-08:00
Duration: 3 min 56 s
Channel: Adi S
Description: A Knudsen cell containing a crystal (M = 157 g mol^-1) is carefully weighed and placed in an evacuated chamber thermostatted at 90 ^∘C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90 ^∘C in Pa from the fact that the weight loss is 39.2 mg.

          A Knudsen cell containing a crystal (M = 157 g mol$^{-1}$) is carefully weighed and placed in an evacuated chamber thermostatted at 90 $^\circ$C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90 $^\circ$C in Pa from the fact that the weight loss is 39.2 mg.

A Knudsen cell containing a crystal (M = 157 g mol^-1) is carefully weighed and placed in an evacuated chamber thermostatted at 90 ^∘C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90 ^∘C in Pa from the fact that the weight loss is 39.2 mg.

Added by Juan L.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Adi S

06/06/2023

Step 1

Given: - Molecular weight of the crystal (M) = 157 g/mol - Weight loss = 39.2 mg = 0.0392 g Number of moles = Weight loss / Molecular weight Number of moles = 0.0392 g / 157 g/mol Number of moles = 0.00025 mol Show more…

Show all steps

Thanks for your feedback!

A Knudsen cell containing a crystal (M = 157 g mol-1) is carefully weighed and placed in an evacuated chamber thermostatted at 90°C for 0.75 hr. The circular hole through which effusion occurs is 0.35 mm in diameter. Calculate the sublimation pressure of the crystal at 90°C in Pa from the fact that the weight loss is 39.2 mg.