how many grams of ZnCl2 are needed to prepare 250 ml of an aqueous solution with an osmolarity equal to 0.300?

An osmolarity of 0.3 osm = 0.1M ZnCl2 (since there are three particles in a molecule of ZnCl2).

You wnat how many mole? That is M x L = ?
The moles ZnCl2 = grams ZnCl2/molar mass.

To determine the number of grams of ZnCl2 needed to prepare the solution, we need to follow these steps:

Step 1: Calculate the molar concentration (Molarity) of the solution.
Given that the osmolarity is 0.300, it is the same as the molar concentration of the solution (M). Therefore, the concentration is 0.300 M.

Step 2: Determine the molar mass of ZnCl2.
The molar mass of ZnCl2 can be calculated by adding the atomic masses of zinc (Zn) and two chlorine (Cl) atoms.
The atomic masses are:
Zn: 65.38 g/mol
Cl: 35.45 g/mol

Molar mass of ZnCl2 = (Zn: 65.38 g/mol) + (Cl: 35.45 g/mol * 2) = 136.28 g/mol

Step 3: Calculate the moles of ZnCl2.
To calculate the moles of ZnCl2, we can use the formula:
moles = molarity * volume (in liters)

Since the volume is given in milliliters (ml), we need to convert it to liters by dividing 250 ml by 1000:
volume (in liters) = 250 ml / 1000 = 0.25 L

moles = 0.300 M * 0.25 L = 0.075 mol

Step 4: Convert moles to grams.
To convert moles to grams, we can use the molar mass of ZnCl2.
grams = moles * molar mass

grams = 0.075 mol * 136.28 g/mol = 10.22 grams

Therefore, you would need approximately 10.22 grams of ZnCl2 to prepare 250 ml of an aqueous solution with an osmolarity equal to 0.300.

To determine the number of grams of ZnCl2 needed to prepare a solution with a specific osmolarity, we need to follow a series of steps.

1. First, we need to calculate the number of moles of solute required using the formula:

Moles of solute = Osmolarity (mol/L) x Volume (L)

In this case, the volume given is in milliliters (ml), so we convert it to liters (L) by dividing by 1000:

Volume (L) = 250 ml / 1000 = 0.25 L

Now we can substitute the given values:

Moles of solute = 0.300 mol/L x 0.25 L = 0.075 moles of solute

2. Next, we need to determine the molecular weight of ZnCl2, which consists of one zinc (Zn) atom and two chlorine (Cl) atoms. The atomic masses of zinc and chlorine are:

Atomic mass of Zn = 65.38 g/mol
Atomic mass of Cl = 35.45 g/mol

To calculate the molecular weight of ZnCl2, we add the atomic masses together:

Molecular weight of ZnCl2 = (1 x atomic mass of Zn) + (2 x atomic mass of Cl)
= (1 x 65.38 g/mol) + (2 x 35.45 g/mol)
= 65.38 g/mol + 70.90 g/mol
= 136.28 g/mol

3. Finally, we can calculate the mass of ZnCl2 using the formula:

Mass (g) = Moles of solute x Molecular weight (g/mol)

Plugging in the values we have:

Mass (g) = 0.075 moles of solute x 136.28 g/mol
= 10.22 grams

Therefore, you would need approximately 10.22 grams of ZnCl2 to prepare 250 ml of an aqueous solution with an osmolarity equal to 0.300.