Molality of the acid produced by the dilution?

Question

Approximately 2.29 mL of concentrated sulfuric acid (specific gravity is 1.84 is added to water. The dilute sulfuric acid is titrated against NaOH and the concentration of H2SO4 is 0.184 mol dm^-3.

How do I calculate the molality of the acid produced by the dilution?

Answer 1

Essentially, you're given the molarity of the dilute acid, from which you can deduce the molality by finding the weight of the water in 1 L of solution.

Step 1: Find out the mass of acid in 1 L of solution.
mass = .184 moles * molecular mass of H2SO4
2 H (1) + 4 O (16) + S (32) = 96 g/mol
.184 mol * 96 g/mol = 17.664 g H2SO4 / L

Step 2: Using the mass and specific gravity, find the volume of acid.
17.664 g / 1.84 g/cm^3 = 9.239 cm^3 of acid

Step 3: Given the volume of acid, and 1 liter of solution, find the volume of water. Since the density of water is 1 g/cm^3, that will also equal the mass in g.
Vw = 1000 cm^3 (1 L) solution - 9.239 cm^3 of acid
Vw = 990.761 cm^3
mw = 990.761 g

Step 4: Find molality given # of moles of H2SO4 and weight of the water in 1 L of solution:
molality = .184 mol / .990761 kg water
molality = 0.182 molal (solution!)

Answer 2

molality of a substance is not any.of moles / molecular mass. image : small "m" therez a formulation corresponding to molarity : m1v1 = m2v2 m=no of moles/molar mass substistute that decrease back interior the formulation i juss gave..however the values given interior the ques. locate the single wich is asked