A chemistry lab experiment requires 34.0 g of silver nitrate. How many grams of a 23.1% by mass solution of silver nitrate should be used?
23.1% by mass means the solution has 23.1 g AgNO3 per 100 grams of solution.
You want 34.0 g; therefore,
100 g soln x (34.0/23.1) = ?? g soln needed.
You can look at it another way; i.e.,
you want (34.0/x g soln) = 0.231 and solve for x.
To determine how many grams of a 23.1% by mass solution of silver nitrate should be used, we need to calculate the amount of silver nitrate present in the solution.
Step 1: Convert the given percentage to a decimal.
The percentage given is 23.1%. To convert it to a decimal, divide it by 100:
23.1% ÷ 100 = 0.231
Step 2: Use the percentage and the desired quantity of silver nitrate to find the required mass.
Let's assume 'x' represents the mass of the 23.1% solution. The equation can be written as:
0.231x = 34.0
To solve for 'x', divide both sides of the equation by 0.231:
x = 34.0 ÷ 0.231
x = 147.17 grams
Therefore, you would need approximately 147.17 grams of a 23.1% by mass solution of silver nitrate for the chemistry lab experiment.