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garri49 [273]
3 years ago
11

0.25 mol of nitric acid is used to make a 0.10 M nitric acid solution. How many grams of solute was used?

Chemistry
1 answer:
Yanka [14]3 years ago
7 0

Answer:

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

Explanation:

Step 1: Data given

Nitric acid = HNO3

Molar mass of H = 1.01 g/mol

Molar mass of N = 14.0 g/mol

Molar mass O = 16.0 g/mol

Number of moles nitric acid (HNO3) = 0.25 moles

Molairty = 0.10 M

Step 2: Calculate molar mass of nitric acid

Molar mass HNO3 = Molar mass H + molar mass N + molar mass (3*O)

Molar mass HNO3 = 1.01 + 14.0 + 3*16.0

Molar mass HNO3 = 63.01 g/mol

Step 3: Calculate mass of solute use

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.25 moles * 63.01 g/mol

Mass HNO3 = 15.75 grams

15.75 grams of HNO3 was used and dissolved in 2.5 liters of solvent, to make a 0.10 M solution

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What is the volume of 9.5 g fluorine gas, F2, at STP?
Tju [1.3M]

Answer:

5.6L

Explanation:

At STP, the pressure and temperature of an ideal gas is

P = 1 atm

T = 273.15k

Volume =?

Mass = 9.5g

From ideal gas equation,

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant =0.082J/mol.K

T = temperature of the ideal gas

Number of moles = mass / molar mass

Molar mass of F2 = 37.99g/mol

Number of moles = mass / molar mass

Number of moles = 9.5 / 37.99

Number of moles = 0.25moles

PV = nRT

V = nRT/ P

V = (0.25 × 0.082 × 273.15) / 1

V = 5.599L = 5.6L

The volume of the gas is 5.6L

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Explanation:

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