342gm of 20% by mass of ba (OH)2 solution sq gr 0.57 is reacted with 1200ml of 2M hn03. If the final density is same as pure water then molarity of the ion in resulting solution by nature of the above solution is dentified

0.888m

Who solve it for me step by step
thank a lot

To solve this problem, we need to follow a step-by-step process:

Step 1: Calculate the moles of barium hydroxide (Ba(OH)2) in the given solution.
Given that the solution has a mass of 342g and a concentration of 20% by mass, we can calculate the mass of Ba(OH)2 in the solution:
Mass of Ba(OH)2 = 20% of 342g = 0.20 * 342g = 68.4g

Next, we need to calculate the moles of Ba(OH)2 using its molar mass. The molar mass of Ba(OH)2 is:
Ba = 1 * atomic mass of Ba = 1 * 137.33 g/mol = 137.33 g/mol
O = 2 * atomic mass of O = 2 * 16.00 g/mol = 32.00 g/mol
H = 2 * atomic mass of H = 2 * 1.01 g/mol = 2.02 g/mol

Molar mass of Ba(OH)2 = 137.33 + 32.00 + 2.02 = 171.35 g/mol

Now, we can calculate the moles of Ba(OH)2 in the solution:
Moles of Ba(OH)2 = Mass of Ba(OH)2 / Molar mass of Ba(OH)2 = 68.4g / 171.35 g/mol = 0.399 mol

Step 2: Calculate the moles of HNO3 in the given solution.
Given that we have 1200 mL of 2M HNO3, we can calculate the moles of HNO3:
Moles of HNO3 = Concentration (M) * Volume (L) = 2M * (1200 mL / 1000 mL/L) = 2.4 mol

Step 3: Determine the limiting reagent.
In a chemical reaction, the limiting reagent is the reactant that is completely consumed and determines the maximum amount of products that can be formed. To determine the limiting reagent, we compare the moles of each reactant.

In the balanced chemical equation:
Ba(OH)2 + 2HNO3 -> Ba(NO3)2 + 2H2O

We can see that the stoichiometric ratio between Ba(OH)2 and HNO3 is 1:2.

By comparing the moles of Ba(OH)2 and HNO3, we see that we have an excess of Ba(OH)2 since 0.399 mol < 2.4 mol. Therefore, Ba(OH)2 is the limiting reagent.

Step 4: Calculate the moles of the product formed.
Since Ba(OH)2 is the limiting reagent, we can use the stoichiometry of the reaction to calculate the moles of the product, Ba(NO3)2.

From the balanced equation, we know that 1 mol of Ba(OH)2 reacts to form 1 mol of Ba(NO3)2:
Moles of Ba(NO3)2 = Moles of Ba(OH)2 = 0.399 mol

Step 5: Calculate the molarity of the ion in the resulting solution.
To find the molarity of the ions in the resulting solution, we need to divide the moles of the ion by the volume of the resulting solution.

Given that the final density is the same as pure water, we can assume that the volume of the resulting solution is equal to the volume of the HNO3 used, which is 1200 mL.

Converting the volume to liters: 1200 mL = 1200 mL / 1000 mL/L = 1.2 L

Molarity of the ion = Moles of Ba(NO3)2 / Volume of solution = 0.399 mol / 1.2 L = 0.333 M (rounded to three decimal places)

Therefore, the molarity of the ion in the resulting solution, by nature of the above solution, is 0.333 M.

Is that m or M (molality or molarity)

M is molarity
m is molality
Did you make a typo? Is that question for molarity of the H^+. That is 0.888M (not m). Total molarity, however, is much more than 0.888M.