Aqueous concentrated nitric acid is 69% hno3 by weight and has a density of 1.42 g/ml. what is the molarity of this solution?

what is the molality of the solution?

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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"