Answer:
The molality of the solution is 35.3 molal
Explanation:
Step 1: Data given
69 w% HNO3
density = 1.42 g/mL
Molar mass of HNO3 = 63.01 g/mol
Step 2: Calculate mass of HNO3
Consider the volume of the solution = 1L =1000 mL
Mass = density * volume
Mass HNO3 = 1.42 g/mL * 1000mL = 1420 grams
if w% = 69% then the mass of this solution is:
0.69 * 1420 = 979.8 grams
Step 3: Calculate number of moles of HNO3
Number of moles = mass / Molar mass
Number of moles HNO3 = 979.8 grams / 63.01 g/mol = 15.55 moles
Step 4: Calculate molarity
Molarity = moles / volume
Molarity = 15.55 moles / 1L = 15.55 M
Step 5: Calculate mass of water
Mass of water = Total mass of HNO3 solution - mass of 69% nitric solution
Mass of water = 1420 - 979.9 = 440.2 grams
Step 6: Calculate molality
Molality = number of moles of HNO3 per 1000 g of water.
Since there are 15.55 moles in 440.2 grams
There are 15.55/440.2 = 0.0353 moles in 1 gram
In 1000 g of water, there are 0.035 *1000 = 35.3 moles
The molality of the solution is 35.3 molal
