Have 250 mL of a 0.80 M solution of silver nitrate AgNO3 dissolved in water. how many grams of AgNO3 does the solution contain
250 mL of 0.8M solution means there are 0.2M of AgNO3
The mole weight of AgNO3 is
107.9 + 14.01 + 3*16.0 = 169.91g
You have .2M, or 33.99g
To calculate the grams of AgNO3 in a solution, you need to use the molarity (M) of the solution and the volume (V) of the solution. Here's the step-by-step process:
Step 1: Convert the volume from milliliters (mL) to liters (L).
Given: 250 mL
Conversion: 1 L = 1000 mL
Calculation: 250 mL ÷ 1000 mL/L = 0.25 L
Step 2: Multiply the molarity (M) by the volume (V) to get the number of moles (n) of AgNO3.
Given: M = 0.80 M, V = 0.25 L
Calculation: n = M × V = 0.80 M × 0.25 L = 0.20 moles AgNO3
Step 3: Determine the molar mass of AgNO3.
AgNO3 molar mass: Ag = 107.87 g/mol + N = 14.01 g/mol + 3 × O = 3 × 16.00 g/mol
Calculation: 107.87 g/mol + 14.01 g/mol + 3 × 16.00 g/mol = 169.87 g/mol
Step 4: Multiply the number of moles (n) by the molar mass (M) to calculate the grams (g) of AgNO3.
Given: n = 0.20 moles AgNO3, M = 169.87 g/mol
Calculation: g = n × M = 0.20 moles × 169.87 g/mol ≈ 33.974 g
Therefore, the 250 mL of the 0.80 M solution of silver nitrate (AgNO3) contains approximately 33.974 grams of AgNO3.