suppose i want to find out how many mL of 1.0M AgNO3 are needed to provide 169.9g of pure AgNO3. what is step one in solving this problem? How many mL of solution are needed?
Mass=Molarity*volume*molmassofAnNO3
solve that for volume.
Step one in solving this problem is to determine the molar mass of AgNO3 (silver nitrate). The molar mass can be calculated by adding up the atomic masses of its constituent elements.
To do this, follow these steps:
1. Look up the atomic masses of each element in AgNO3:
- Ag (silver): 107.87 g/mol
- N (nitrogen): 14.01 g/mol
- O (oxygen): 16.00 g/mol
2. Multiply the atomic mass of each element by the number of atoms of that element in one molecule of AgNO3:
- Ag: 1 atom x 107.87 g/mol = 107.87 g/mol
- N: 1 atom x 14.01 g/mol = 14.01 g/mol
- O: 3 atoms x 16.00 g/mol = 48.00 g/mol
3. Add up the mass contributions from each element:
- 107.87 g/mol + 14.01 g/mol + 48.00 g/mol = 169.88 g/mol (rounded to two decimal places)
The molar mass of AgNO3 is approximately 169.88 g/mol.
To find out how many mL of 1.0M AgNO3 solution are needed to provide 169.9g of pure AgNO3, further steps would be required.