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ella [17]
2 years ago
5

A 25.0-milliliter sample of HNO3(aq) is neutralized by 32.1 milliliters of 0.150 M KOH(aq). What is the molarity of the HNO3(aq)

?
(1) 0.117 M (3) 0.193 M
(2) 0.150 M (4) 0.300 M
Chemistry
2 answers:
dusya [7]2 years ago
4 0
The answer is (3) 0.193M. To find the molarity of HNO3, you just need to use the M1V1=M2V2 equation (no need to worry about ionization constants because HNO3 is monoprotic and KOH dissociates 1:1). Since the molarity you are looking for is M1, you get M1=M2V2/V1=(0.150)(32.1)/25.0= 0.193M
lubasha [3.4K]2 years ago
3 0

<u>Answer:</u> The correct answer is Option 3.

<u>Explanation:</u>

To calculate the molarity of the acid, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of acid

M_2\text{ and }V_2 are the molarity and volume of base

We are given:

M_1=?M\\V_1=25mL\\M_2=0.150M\\V_2=32.1mL

Putting values in above equation, we get:

M_1\times 25=0.15\times 32.1\\\\M_1=0.193M

Hence, the correct answer is Option 3.

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Let's go over the given information. We have the volume, temperature and pressure. From the ideal gas equation, that's 4 out of 5 knowns. So, we actually don't need Pvap of water anymore. Assuming ideal gas, the solution is as follows:

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The molar mass of H₂ is 2 g/mol.

Mass of H₂ = 7.897×10⁻³ mol * 2 g/mol = <em>0.016 g H₂</em>
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