I think that the answer is d
Answer:
2.52 g NaCl
Explanation:
(Step 1)
To find the mass, you first need to find the moles NaCl. This value can be found using the molarity ratio:
Molarity = moles / volume (L)
After you convert mL to L, you can plug the given values into the equation and simplify to find moles.
136.9 mL / 1,000 = 0.1369 L
Molarity = moles / volume
0.315 M = moles / 0.1369 L
0.0431 = moles
(Step 2)
Now, you can use the molar mass to convert moles to grams.
Molar Mass (NaCl): 22.990 g/mol + 35.453 g/mol
Molar Mass (NaCl): 58.443 g/mol
0.0431 moles NaCl 58.443 g
------------------------------ x ------------------- = 2.52 g NaCl
1 mole
Answer:
0.35 atm
Explanation:
It seems the question is incomplete. But an internet search shows me these values for the question:
" At a certain temperature the vapor pressure of pure thiophene (C₄H₄S) is measured to be 0.60 atm. Suppose a solution is prepared by mixing 137. g of thiophene and 111. g of heptane (C₇H₁₆). Calculate the partial pressure of thiophene vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal."
Keep in mind that if the values in your question are different, your answer will be different too. <em>However the methodology will remain the same.</em>
First we <u>calculate the moles of thiophene and heptane</u>, using their molar mass:
- 137 g thiophene ÷ 84.14 g/mol = 1.63 moles thiophene
- 111 g heptane ÷ 100 g/mol = 1.11 moles heptane
Total number of moles = 1.63 + 1.11 = 2.74 moles
The<u> mole fraction of thiophene</u> is:
Finally, the <u>partial pressure of thiophene vapor is</u>:
Partial pressure = Mole Fraction * Vapor pressure of Pure Thiophene
- Partial Pressure = 0.59 * 0.60 atm
Answer:
Yes Concurred,but where is the question
