how many g of agno3 would you need to make a 575ml solution of 0.25 m agno3
If you expect to get along in chemistry you should find the caps key and use it. There is a difference between 0.25m and 0.25M. I suspect you mean 0.25M.
How many mols do you need? That's M x L = ?
Then mols = grams/molar mass. You know molar mass and mols, solve for grams.
To find out how many grams of AgNO3 would be needed to make a 575 ml solution of 0.25 M AgNO3, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) x volume of solution (in liters)
Convert the volume of solution from milliliters to liters:
575 ml = 575 / 1000 = 0.575 L
Now substitute the given values into the formula:
moles of solute = 0.25 M x 0.575 L
moles of solute = 0.14375 moles (rounded to 5 decimal places)
Finally, calculate the grams of AgNO3 needed using the molar mass of AgNO3:
Molar mass of AgNO3 = 107.87 g/mol (Ag = 107.87, N = 14.01, O = 16.00)
grams of AgNO3 = moles of solute x molar mass of AgNO3
grams of AgNO3 = 0.14375 moles x 107.87 g/mol
grams of AgNO3 ≈ 15.49 g (rounded to 2 decimal places)
Therefore, you would need approximately 15.49 grams of AgNO3 to make a 575 ml solution of 0.25 M AgNO3.