To find the volume of a 0.1 M HCl solution containing 1.46 grams of HCl, we can use the formula:
moles = mass / molar mass
First, we need to calculate the number of moles of HCl:
moles = 1.46 g / 36.46 g/mol = 0.04 moles
Now, we can calculate the volume of the solution using the formula:
Molarity = moles / volume (in liters)
0.1 M = 0.04 moles / volume
Volume = 0.04 moles / 0.1 M = 0.4 liters or 400 milliliters
Therefore, the volume of the 0.1 M HCl solution containing 1.46 grams of HCl is 400 milliliters.
What is the volume of a 0.1 M HCL solution containing one 1.46GM of HCL?
3 answers
What is the volume of a 0.2 M AgNO3 solution containing 8.5 grams of AgNO3
To find the volume of a 0.2 M AgNO3 solution containing 8.5 grams of AgNO3, we can use the formula:
moles = mass / molar mass
First, we need to calculate the number of moles of AgNO3:
molar mass of AgNO3 = 107.87 + 14.01 + 3(16.00) = 169.87 g/mol
moles = 8.5 g / 169.87 g/mol = 0.05 moles
Now, we can calculate the volume of the solution using the formula:
Molarity = moles / volume (in liters)
0.2 M = 0.05 moles / volume
Volume = 0.05 moles / 0.2 M = 0.25 liters or 250 milliliters
Therefore, the volume of the 0.2 M AgNO3 solution containing 8.5 grams of AgNO3 is 250 milliliters.
moles = mass / molar mass
First, we need to calculate the number of moles of AgNO3:
molar mass of AgNO3 = 107.87 + 14.01 + 3(16.00) = 169.87 g/mol
moles = 8.5 g / 169.87 g/mol = 0.05 moles
Now, we can calculate the volume of the solution using the formula:
Molarity = moles / volume (in liters)
0.2 M = 0.05 moles / volume
Volume = 0.05 moles / 0.2 M = 0.25 liters or 250 milliliters
Therefore, the volume of the 0.2 M AgNO3 solution containing 8.5 grams of AgNO3 is 250 milliliters.