what mass of silver(1)nitrate is needed to prepare 100cm3 of silver(1)nitrate solution,concentration 0.2mol/dm3?
the mass of one mole of silver nitrate is 170g
please help me with this question
find the moles of silver nitrate by multiplying conc and vol of same units
then find mass from "moles = mass in grams (divided by) relative molecular mass "
To determine the mass of silver nitrate needed to prepare the desired solution, we can use the formula:
Mass = Volume x Concentration x Molar mass
Given:
Volume = 100 cm3 (which is equivalent to 100/1000 = 0.1 dm3)
Concentration = 0.2 mol/dm3
Molar mass of silver nitrate = 170 g/mol
Substituting the values into the formula:
Mass = 0.1 dm3 x 0.2 mol/dm3 x 170 g/mol
Mass = 0.02 mol x 170 g/mol
Mass = 3.4 g
Therefore, you would need 3.4 grams of silver nitrate to prepare 100 cm3 of a silver nitrate solution with a concentration of 0.2 mol/dm3.
To find the mass of silver nitrate needed to prepare the solution, we can use the equation:
concentration (mol/dm^3) = (number of moles) / (volume in dm^3)
Given:
- Concentration = 0.2 mol/dm^3
- Volume = 100 cm^3
First, we need to convert the volume from cm^3 to dm^3 by dividing it by 1000:
Volume = 100 cm^3 / 1000 = 0.1 dm^3
Now we can rearrange the equation to find the number of moles:
(number of moles) = concentration (mol/dm^3) x volume (dm^3)
(number of moles) = 0.2 mol/dm^3 x 0.1 dm^3
(number of moles) = 0.02 moles
The molar mass of silver nitrate is given as 170 g/mol. We can use this to calculate the mass of silver nitrate needed:
mass = (number of moles) x molar mass
mass = 0.02 moles x 170 g/mol
mass = 3.4 grams
Therefore, you would need approximately 3.4 grams of silver nitrate to prepare 100 cm^3 of a solution with a concentration of 0.2 mol/dm^3.
so, how many moles of AgNO3 are needed?
multiply that by 170 g/mol