154.99 moles of HNO3 get dissolved in 100ml of water when 10.0 ml of conc. HNO3 (15.5M) is added to it.
<h3>How does HNO3 dissolve in water ?</h3>
Since nitric acid is a moderately potent acid, we assume that the water solvent undergoes complete protonolysis in accordance with the equation:
⇄ ![H_{3}O^{+} + NO_{3}^{-}](https://tex.z-dn.net/?f=H_%7B3%7DO%5E%7B%2B%7D%20%20%2B%20%20NO_%7B3%7D%5E%7B-%7D)
To the right is where the equilibrium is.
The aqueous solution is now stoichiometric for hydronium ion and nitrate ion, regardless of the initial nitric acid concentration.
<h3>Given: </h3>
V1 (Volume of nitric acid) = 10.0ml
M1 (Molarity of nitric acid) = 15.5M
V (Volume of Water) = 100.0ml
n (no. of moles of HNO3 dissolved) = ?
<h3>Formula applied:</h3>
1) For dilution : M1V1 = M2V2
where : V2 = total volume of the final solution
M2 = molarity of the final solution
2) Molarity = ( no. of moles (mol) ) / ( total volume of the solution (l) )
<h3 /><h3>Universal constants used :</h3>
Molar mass of HNO3 = 63.01g
<h3 /><h3>Solution:
</h3>
![M1*V1 = M2*V2 = > M2 = \frac{M1*V1}{V2} \\\\= > M2 = \frac{15.5*10}{110} mol ml^{-1} \\\\= > M2 = 1.409 mol ml^{-1}\\](https://tex.z-dn.net/?f=M1%2AV1%20%3D%20M2%2AV2%20%20%20%20%3D%20%3E%20M2%20%3D%20%20%5Cfrac%7BM1%2AV1%7D%7BV2%7D%20%5C%5C%5C%5C%3D%20%3E%20M2%20%3D%20%5Cfrac%7B15.5%2A10%7D%7B110%7D%20mol%20ml%5E%7B-1%7D%20%5C%5C%5C%5C%3D%20%3E%20M2%20%3D%201.409%20mol%20ml%5E%7B-1%7D%5C%5C)
Now, calculate the no. of moles dissolved in the solution :
![M2 = \frac{n}{V} \\\\= > 1.409 mol ml^{-1} = \frac{n}{110} ml \\\\= > n = 1.409 * 110 mol\\\\= > n = 154.99 mol](https://tex.z-dn.net/?f=M2%20%3D%20%5Cfrac%7Bn%7D%7BV%7D%20%5C%5C%5C%5C%3D%20%3E%201.409%20mol%20ml%5E%7B-1%7D%20%20%3D%20%20%5Cfrac%7Bn%7D%7B110%7D%20ml%20%5C%5C%5C%5C%3D%20%3E%20n%20%20%3D%20%201.409%20%2A%20110%20mol%5C%5C%5C%5C%3D%20%3E%20n%20%3D%20154.99%20mol)
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