How to find the %(W/V) of 134mL of a 0.610 M solution of Mg(OH)2

0.610M means 0.610 moles/L soln or 0.0610 moles/100 mL.

mol Mg(OH)2 = grams Mg(OH)2/molar mass. You know molar mass and moles, solve for grams. That many grams/100 mL is the w/v %.

To find the %(W/V) of a solution, you need to know the mass of the solute (in grams) and the volume of the solution (in mL). In this case, we are given a volume of solution but not the mass of Mg(OH)2.

To find the mass of Mg(OH)2, we'll use the relationship between moles, molarity, and volume.

1. Convert the volume of the solution from mL to L:
134 mL = 134/1000 L = 0.134 L

2. Use the molarity and volume to find the moles of Mg(OH)2:
moles = molarity × volume
moles = 0.610 M × 0.134 L = 0.08174 moles

3. Determine the molar mass of Mg(OH)2:
Mg: 24.31 g/mol
O: 16.00 g/mol
H: 1.01 g/mol (2 hydrogen atoms)
Total molar mass = (24.31 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol)
= 58.33 g/mol

4. Calculate the mass of Mg(OH)2 in grams:
mass = moles × molar mass
mass = 0.08174 moles × 58.33 g/mol = 4.767 grams

Now that we have the mass of Mg(OH)2 (4.767 grams) and the volume of the solution (134 mL), we can calculate the %(W/V) using the following formula:

%(W/V) = (mass of solute / volume of solution) × 100

%(W/V) = (4.767 g / 134 mL) × 100 = 3.55% (rounded to two decimal places)

Therefore, the %(W/V) of the 134 mL solution of 0.610 M Mg(OH)2 is approximately 3.55%.

To find the %(W/V) of a solution, you need to use the formula:

%(W/V) = (mass of solute / volume of solution) * 100

Here's how you can find the %(W/V) of the given solution of Mg(OH)2:

Step 1: Calculate the mass of solute.
To do this, you need to know the molar mass of Mg(OH)2. The molar mass of Mg(OH)2 is calculated as follows:
Mg: 24.31 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
So, the molar mass of Mg(OH)2 is: 24.31 + (16.00 + 1.01*2) = 58.33 g/mol

Now, use the molar mass to calculate the mass of Mg(OH)2 in the solution.
mass of Mg(OH)2 = molar mass * moles
moles = concentration (M) * volume (L)
volume (L) = volume (mL) / 1000
mass of Mg(OH)2 = 58.33 g/mol * 0.610 mol/L * 134 mL / 1000 mL = 4.161 g

Step 2: Calculate the volume of the solution.
Given: volume of solution = 134 mL

Step 3: Plug the values into the formula.
%(W/V) = (mass of solute / volume of solution) * 100
%(W/V) = (4.161 g / 134 mL) * 100 = 3.10 % (rounded to two decimal places)

Therefore, the %(W/V) of the 0.610 M solution of Mg(OH)2 is approximately 3.10%.