Determine the grams of solute required to prepare 0.5 m solution HC2H3O containing 750 g of water. Make sure to show your work.
**(Thanks for helping by the way.)**
.5=molessolute/.750
molessolute=mass/molmass
mass=.5*.750*molmassHC2H3O
To determine the grams of solute required to prepare a 0.5 M solution of HC2H3O, we need to know the molar mass of HC2H3O.
The molar mass of HC2H3O can be calculated by adding up the atomic masses of the elements in its chemical formula:
Molar mass of H (hydrogen) = 1 g/mol
Molar mass of C (carbon) = 12 g/mol
Molar mass of O (oxygen) = 16 g/mol
There are 2 hydrogen atoms, 2 carbon atoms, and 3 oxygen atoms in one molecule of HC2H3O. So, the molar mass of HC2H3O can be calculated as follows:
Molar mass of HC2H3O = (2 * Molar mass of H) + (2 * Molar mass of C) + (3 * Molar mass of O)
= (2 * 1 g/mol) + (2 * 12 g/mol) + (3 * 16 g/mol)
= 2 g/mol + 24 g/mol + 48 g/mol
= 74 g/mol
Now, let's calculate the number of moles of HC2H3O required to prepare a 0.5 M solution.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, to calculate the moles of solute required, we can use the formula:
Moles of solute = Molarity * Volume (in liters)
In this case, the molarity is 0.5 M and the volume is not given. However, we know that the solution contains 750 g of water.
To convert the mass of water (in grams) to volume (in liters), we need to know the density of water. The density of water at room temperature is approximately 1 g/mL or 1 g/cm³.
Therefore, the volume of water can be calculated as:
Volume of water (in liters) = Mass of water (in grams) / Density of water (in g/cm³)
In this case, the mass of water is 750 g and the density of water is 1 g/cm³. Therefore:
Volume of water (in liters) = 750 g / 1 g/cm³
= 750 cm³ / 1000 cm³/mL
= 0.75 L
Now, let's calculate the moles of solute required:
Moles of solute = Molarity * Volume
= 0.5 M * 0.75 L
= 0.375 mol
Finally, we can calculate the grams of solute required:
Grams of solute = Moles of solute * Molar mass
= 0.375 mol * 74 g/mol
= 27.75 g
Therefore, you would need approximately 27.75 grams of HC2H3O to prepare a 0.5 M solution using 750 g of water.