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Answer : Both solutions contain molecules.
Explanation : The number of molecules of 0.5 M of sucrose is equal to the number of molecules in 0.5 M of glucose. Both solutions contain molecules.
Avogadro's Number is =
which represents particles per mole and particles may be typically molecules, atoms, ions, electrons, etc.
Here, only molarity values are given; where molarity is a measurement of concentration in terms of moles of the solute per liter of solvent.
Since each substance has the same concentration, 0.5 M, each will have the same number of molecules present per liter of solution.
Addition of molar mass for individual substance is not needed. As if both are considered in 1 Liter they would have same moles which is 0.5.
We can calculate the number of molecules for each;
Number of molecules = ;
∴ Number of molecules = which will be =
Thus, these solutions compare to each other in that they have not only the same concentration, but they will have the same number of solvated sugar molecules. But the mass of glucose dissolved will be less than the mass of sucrose.
Answer:
0.1035 M
Explanation:
Considering:
Sodium chloride will furnish Sodium ions as:
Given :
For Sodium chloride :
Molarity = 0.288 M
Volume = 3.58 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 3.58×10⁻³ L
Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:
Moles of sodium ions by sodium chloride = 0.00103104 moles
Sodium sulfate will furnish Sodium ions as:
Given :
For Sodium sulfate :
Molarity = 0.001 M
Volume = 6.51 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.51 ×10⁻³ L
Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:
Moles of sodium ions by Sodium sulfate = 0.00001302 moles
Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles
Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L
Concentration of sodium ions is:
<u> The final concentration of sodium anion = 0.1035 M</u>