5,10g of pure silver nitrate is dessolved in 250cm of water .calculate the concentration of solution.(0,120mol.dm)
To calculate the concentration of the solution, we need to convert the mass of silver nitrate to moles using its molar mass. The molar mass of AgNO3 is 169.87 g/mol.
Number of moles of AgNO3 = mass of AgNO3 / molar mass of AgNO3
Number of moles of AgNO3 = 5.10 g / 169.87 g/mol = 0.0300 mol
Now, we need to calculate the volume of the solution in dm^3. We convert 250 cm^3 to dm^3 by dividing by 1000.
Volume of solution = 250 cm^3 / 1000 = 0.250 dm^3
Finally, we can calculate the concentration of the solution by dividing the number of moles of AgNO3 by the volume of the solution.
Concentration of solution = Number of moles of AgNO3 / Volume of solution
Concentration of solution = 0.0300 mol / 0.250 dm^3 = 0.120 mol/dm3
Therefore, the concentration of the solution is 0.120 mol/dm3.
To calculate the concentration of the solution, we need to convert the given information into the correct units.
1. Convert the given mass of silver nitrate from grams to moles:
The molar mass of silver nitrate (AgNO3) is:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (three oxygen atoms)
Total molar mass of AgNO3:
107.87 + 14.01 + (16.00 * 3) = 169.87 g/mol
Now, we can calculate the number of moles of silver nitrate:
Moles of AgNO3 = Mass of AgNO3 / Molar mass of AgNO3
Moles of AgNO3 = 5.10 g / 169.87 g/mol = 0.0300 mol
2. Convert the given volume of water from cm³ to dm³:
1 dm³ = 1000 cm³
Volume of water = 250 cm³ / 1000 cm³/dm³ = 0.250 dm³
3. Calculate the concentration of the solution:
Concentration (mol/dm³) = Moles of solute / Volume of solution (in dm³)
Concentration = 0.0300 mol / 0.250 dm³ = 0.120 mol/dm³
Therefore, the concentration of the solution is 0.120 mol/dm³.