How many kilograms of water must be added to 75.5 grams of Calcium nitrate to form a 0.500 m solution?
see other post.
To find out how many kilograms of water must be added to 75.5 grams of Calcium nitrate to form a 0.500 m solution, we can use the concept of molarity.
First, let's determine the number of moles of Calcium nitrate in 75.5 grams. To do this, we need to divide the mass of Calcium nitrate by its molar mass. The molar mass of Calcium nitrate (Ca(NO3)2) is 164.1 g/mol.
Number of moles of Calcium nitrate = Mass / Molar mass
Number of moles of Calcium nitrate = 75.5 g / 164.1 g/mol
Now, let's calculate the volume of the solution needed to achieve a concentration of 0.500 m. The formula for molarity (m) is defined as moles of solute divided by liters of solution.
Molarity (m) = Moles of solute / Volume of solution (in liters)
We can rearrange the formula to solve for the volume of the solution (in liters):
Volume of solution (in liters) = Moles of solute / Molarity
In this case, the molarity is given as 0.500 m, and the moles of solute are the number of moles of Calcium nitrate we calculated earlier.
Volume of solution (in liters) = (75.5 g / 164.1 g/mol) / 0.500 mol/L
Now, to find the mass of water required to achieve the desired volume, we can use the fact that the density of water is approximately 1 g/mL, or 1 kg/L.
Mass of water = Volume of solution (in liters) * Density of water
Mass of water = Volume of solution (in liters) * 1 kg/L
Finally, plug in the values we calculated earlier to find the mass of water needed:
Mass of water = [(75.5 g / 164.1 g/mol) / 0.500 mol/L] * 1 kg/L
Calculating this equation will give you the mass of water required in kilograms to form a 0.500 m Calcium nitrate solution.