Silver Nitrate and Sodium carbonate has a known concentration of 0.500M. preparing 25ml of 0.500m. Calculate the mass to the nearest 0.01g.
Your problem isn't clear. You want to prepare two solutions of that strength? Is that 0.500m or 0.500M
To calculate the mass of the solid compound (either silver nitrate or sodium carbonate) needed to prepare a 25 mL solution of 0.500 M concentration, we will use the equation:
M1V1 = M2V2
Where:
M1 and V1 are the initial concentration and volume, and
M2 and V2 are the final concentration and volume.
Let's consider silver nitrate first:
M1 = 0.500 M (given)
V1 = unknown
M2 = 0.500 M (given)
V2 = 25 mL
Using the equation, we can rearrange it to solve for V1:
V1 = (M2 * V2) / M1
V1 = (0.500 M * 25 mL) / 0.500 M
V1 = 25 mL
So, to prepare a 25 mL solution of 0.500 M silver nitrate, you need 25 mL of silver nitrate solution.
Now, let's calculate the mass to the nearest 0.01 g. The molar mass of silver nitrate (AgNO3) is 169.87 g/mol.
Mass of AgNO3 = Molar mass * Volume
Mass of AgNO3 = 169.87 g/mol * 0.025 L (since 25 mL = 0.025 L)
Now, calculate the mass:
Mass of AgNO3 = 169.87 g/mol * 0.025 L
Mass of AgNO3 = 4.2475 g
Rounded to the nearest 0.01 g, the mass of silver nitrate needed is 4.25 grams.
You can follow the same steps to calculate the mass of sodium carbonate (Na2CO3) using its molar mass, which is 105.99 g/mol.