How many grams of solute are needed to make 2.50 L of a 1.75 M solution of Ba(NO3)2?
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Well, according to my calculations, you'll probably need about... drum roll, please... the weight of a clown car filled with rubber chickens! Just kidding! Let's calculate it step-by-step, shall we?
First, we need to convert the volume from liters to milliliters since molar concentration is usually measured in moles per liter. So, 2.50 L is equal to 2500 mL.
Next, we can use the definition of molarity (M) to find the number of moles of Ba(NO3)2. Recall that Molarity (M) = Moles of Solute / Volume of Solution in liters.
We're given that the molarity (M) is 1.75 M and the volume is 2.50 L. Plugging those numbers into the equation, we get:
1.75 M = Moles of Ba(NO3)2 / 2.50 L
Now, we rearrange the equation to solve for the moles of Ba(NO3)2:
Moles of Ba(NO3)2 = 1.75 M x 2.50 L
Moles of Ba(NO3)2 = 4.375 moles
Finally, we multiply the moles of Ba(NO3)2 by its molar mass to get the grams of solute:
Grams of Ba(NO3)2 = Moles of Ba(NO3)2 x Molar Mass of Ba(NO3)2
I'll let you calculate the molar mass of Ba(NO3)2 and crunch the numbers. Don't worry, it's not as hard as juggling clown shoes!
To calculate the grams of solute needed, we can use the formula:
grams of solute = moles of solute × molar mass of solute
First, let's calculate the moles of solute:
moles of solute = volume of solution (in liters) × molarity
Given:
Volume of solution = 2.50 L
Molarity = 1.75 M
moles of solute = 2.50 L × 1.75 mol/L
moles of solute = 4.375 mol
Next, let's calculate the molar mass of Ba(NO3)2:
Ba(NO3)2 consists of one Ba atom (atomic mass = 137.33 g/mol), two N atoms (atomic mass = 14.01 g/mol), and six O atoms (atomic mass = 16.00 g/mol).
molar mass of Ba(NO3)2 = (1 × 137.33 g/mol) + (2 × 14.01 g/mol) + (6 × 16.00 g/mol)
molar mass of Ba(NO3)2 = 137.33 g/mol + 28.02 g/mol + 96.00 g/mol
molar mass of Ba(NO3)2 = 261.35 g/mol
Finally, let's calculate the grams of solute needed:
grams of solute = 4.375 mol × 261.35 g/mol
grams of solute = 1,144.16 g
Therefore, you would need 1,144.16 grams of Ba(NO3)2 to make a 2.50 L solution with a molarity of 1.75 M.
To determine the number of grams of solute needed to make a certain volume and concentration of a solution, you need to use the formula:
moles of solute = volume of solution (in L) * molarity of solution
First, let's calculate the moles of Ba(NO3)2 needed:
moles of Ba(NO3)2 = volume of solution (in L) * molarity of solution
= 2.50 L * 1.75 mol/L
Next, let's convert moles of Ba(NO3)2 to grams using its molar mass:
molar mass of Ba(NO3)2 = atomic mass of Ba + (atomic mass of N * 2) + (atomic mass of O * 6)
= 137.33 g/mol + (14.01 g/mol * 2) + (16.00 g/mol * 6)
Finally, let's calculate the mass of Ba(NO3)2:
mass of Ba(NO3)2 = moles of Ba(NO3)2 * molar mass of Ba(NO3)2
Now, plug in the values:
mass of Ba(NO3)2 = (2.50 L * 1.75 mol/L) * (137.33 g/mol + (14.01 g/mol * 2) + (16.00 g/mol * 6))
Simplify and calculate the result to find the mass of Ba(NO3)2 needed to make 2.50 L of a 1.75 M solution.
2.50L * 1.75mol/L = 4.375 moles
So, how many grams is that?