342gm of 20% by mass of ba(oh)2 solution sq gr .57 is reacted with 1200ml of 2m hn03. if the find density is same as pure water then molarity of the ion in resulting solution by nature of the above solution is dentified.

Question

342gm of 20% by mass of ba(oh)2 solution sq gr .57 is reacted with 1200ml of 2m hn03. if the find density is same as pure water then molarity of the ion in resulting solution by nature of the above solution is dentified.

The answer 0.888M

Who helps me to solve it step step

Thank a lot all tutors

Answer 1

Is that 2m HNO3 or 2M HNO3?

Answer 2

To solve this problem step by step, you need to use the concept of stoichiometry and molarity.

Step 1: Calculate the moles of Ba(OH)2 in the given solution.
To do this, you need to know the molar mass of Ba(OH)2.
Molar mass of Ba(OH)2 = (137.33 g/mol + 2 * 16.00 g/mol + 2 * 1.01 g/mol) = 171.33 g/mol

Now, calculate the moles of Ba(OH)2:
Moles of Ba(OH)2 = mass / molar mass
= 342 g / 171.33 g/mol
= 2 moles

Step 2: Calculate the moles of HNO3 in the given solution.
To do this, you need to use the formula: Moles = Molarity * Volume (in liters)
Molarity of HNO3 = 2 M
Volume of HNO3 = 1200 mL = 1200 cm^3 = 1200/1000 L = 1.2 L

Moles of HNO3 = 2 M * 1.2 L
= 2.4 moles

Step 3: Identify the reactant in excess.
This can be done by comparing the stoichiometric ratio between the two compounds. The balanced equation is:

Ba(OH)2 + 2 HNO3 -> Ba(NO3)2 + 2 H2O

From the balanced equation, 1 mole of Ba(OH)2 reacts with 2 moles of HNO3.

Since you have 2 moles of Ba(OH)2 and 2.4 moles of HNO3, you can see that HNO3 is in excess.

Step 4: Calculate the moles of Ba(NO3)2 formed.
From the balanced equation, 1 mole of Ba(OH)2 produces 1 mole of Ba(NO3)2.

Moles of Ba(NO3)2 = 2 moles of Ba(OH)2 (since HNO3 is in excess)

Step 5: Calculate the molarity of Ba(NO3)2 in the resulting solution.
To do this, you need to know the volume of the resulting solution.
From the given conditions, it states the density is the same as pure water. Therefore, we can assume the volume of the solution is equal to the volume of the HNO3 solution added (1.2 L).

Molarity of Ba(NO3)2 = Moles of Ba(NO3)2 / Volume of solution (in liters)
= 2 moles / 1.2 L
= 1.67 M

However, the question asks for the molarity of the ion, not the entire salt Ba(NO3)2. Since Ba(NO3)2 dissociates into 3 ions (1 Ba2+ ion and 2 NO3- ions), you need to multiply the molarity by the number of ions.

Molarity of the ion = Molarity of Ba(NO3)2 * Number of ions
= 1.67 M * 3
= 5.01 M

Therefore, the molarity of the ion in the resulting solution is identified as 5.01 M.