Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
I am going to say it is false.
Answer:
Option D. pH= 1.3 strong acid
Explanation:
From the question given:
The hydrogen ion concentration [H+] = 0.05 M
pH = —Log [H+]
pH = —Log 0.05
pH = 1.3
Since the pH lies between 0 and 7, the solution is acidic. Since the pH value is low, the solution is a strong acid
The answer to this question would be: lower molar concentration
Osmotic pressure is influenced by the number of ions and the concentration of the molecule in the solution. In NaCl, the molecule will split into 1 Na+ ion and 1 Cl- ion which results in 2 ions per compound. In MgCl2, the compound will split into 1 Mg2+ ion and 2 Cl- ion which results in 3 ions. Therefore, the osmotic pressure of MgCl2 will be 3/2 times of NaCl.
MgCl2 will need less concentration to achieve same osmotic pressure as NaCl. If the MgCl2 solution is isotonic with NaCl, the concentration of MgCl2 would be lower than NaCl
1 ounces ----------- 29.5735 mL
5 ounces ----------- ??
5 x 29.5735 / 1 => 147.868 mL
hope this helps!
