A 3.87 g sample of a hydrate of MnCl2 was heated to drive off the water. The anhydrous powder weighed 2.46 g. Determine the amount of water lost and calculate the number of moles of water there are to every one mole of the hydrate.
A 3.87 g sample of a hydrate of MnCl2 was heated to drive off the water. The anhydrous powder weighed 2.46 g. Determine the amount of water lost and calculate the number of moles of water there are to every one mole of the hydrate.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question number 2.
![**Question 1:**
A certain hydrate salt that has a molar mass of 474 g/mole consists of 45.6% water. Calculate the number of water moles there are to every one mole of the salt.
**Solution:**
1. Determine the mass of water:
\[
\frac{45.6}{100} \times 474 \, \text{g/mol} = 216.144 \, \text{g/mol}
\]
2. Molar mass of water = 18 g/mol.
3. Calculate the number of moles of water:
\[
\frac{216.144 \, \text{g/mol}}{18 \, \text{g/mol}} = 12
\]
**Question 2:**
A 3.87 g sample of a hydrate of MnCl₂ was heated to drive off the water. The anhydrous powder weighed 2.46 g. Determine the amount of water lost and calculate the number of moles of water there are to every one mole of the hydrate.
_No further solution provided._
**Explanation:**
In the solved question (1), the hydrate salt contains water amounting to 45.6% of its mass. The steps involve calculating the actual mass of water within one mole of the hydrate using the percentage given, and then dividing by the molar mass of water to find the number of moles of water per mole of salt. The number of moles calculated is 12.
Question 2 outlines a similar exercise but lacks detailed work, presumably requiring the calculation of moles of water based on the mass difference pre- and post-heating.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd19f3b21-fb4f-4e37-bd58-40b9ca0f2cef%2F8aa023c6-68a0-4f0b-9e24-166bcd910582%2Fwwc69sb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 1:**
A certain hydrate salt that has a molar mass of 474 g/mole consists of 45.6% water. Calculate the number of water moles there are to every one mole of the salt.
**Solution:**
1. Determine the mass of water:
\[
\frac{45.6}{100} \times 474 \, \text{g/mol} = 216.144 \, \text{g/mol}
\]
2. Molar mass of water = 18 g/mol.
3. Calculate the number of moles of water:
\[
\frac{216.144 \, \text{g/mol}}{18 \, \text{g/mol}} = 12
\]
**Question 2:**
A 3.87 g sample of a hydrate of MnCl₂ was heated to drive off the water. The anhydrous powder weighed 2.46 g. Determine the amount of water lost and calculate the number of moles of water there are to every one mole of the hydrate.
_No further solution provided._
**Explanation:**
In the solved question (1), the hydrate salt contains water amounting to 45.6% of its mass. The steps involve calculating the actual mass of water within one mole of the hydrate using the percentage given, and then dividing by the molar mass of water to find the number of moles of water per mole of salt. The number of moles calculated is 12.
Question 2 outlines a similar exercise but lacks detailed work, presumably requiring the calculation of moles of water based on the mass difference pre- and post-heating.
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