Calculate the volume of a 36% solution of hydrochloric acid (density = 1.50g/mL, molar mass = 36 g/mol) required to prepare 9 liters of a 5 molar solution.
A. 1 liter
B. 2 liters
C. 3 liters
D. 4 liters
E. 5 liters
To prepare 9 litres of 5 molar we need
45 moles of HCl
which is 36x45 g = 1620 g of HCl
if the stock solution is 36w/w% then we need
1620 g/0.36 of the stock = 4500 g of the stock.
to find the volume needed
volume = mass/density
volume = 4500 g/1.50 g/ml
=3000 ml
To calculate the volume of the 36% solution of hydrochloric acid required, we can follow these steps:
1. Start by determining the mass of hydrochloric acid needed to prepare the 9 liters of a 5 molar solution. The molar mass of hydrochloric acid is given as 36 g/mol. Since we want a 5 molar solution, we need 5 moles of hydrochloric acid per liter.
Mass of hydrochloric acid = molar mass × moles
Mass of hydrochloric acid = 36 g/mol × 5 mol/L × 9 L
Mass of hydrochloric acid = 1,620 g
2. Next, determine the volume of the 36% solution needed to provide the required mass of hydrochloric acid. Since the density of the 36% solution is given as 1.50 g/mL, we can use the formula:
Volume = Mass / Density
Volume = 1,620 g / 1.50 g/mL
Volume ≈ 1,080 mL
3. Convert the volume from milliliters to liters:
Volume = 1,080 mL / 1,000 mL/L
Volume = 1.08 liters
Therefore, the volume of the 36% solution of hydrochloric acid required to prepare 9 liters of a 5 molar solution is approximately 1.08 liters.
The closest option from the given choices is:
A. 1 liter.