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3 years ago

I’ve gotten the answer as x=.478 but when plugging back in, .46/.478 ≠ .22. Help please

Chemistry

1 answer:

ivanzaharov [21]3 years ago

6 0

Answer:

15. 2.66 moles .

16. 2.09L.

Explanation:

Molarity of a solution is simply defined as the mole of solute per unit litre of the solvent. Mathematically, it is represented as:

Molarity = mole /Volume.

With the above formula, let us answer the questions given above

15. Data obtained from the question include the following:

Volume of solution = 1.4L

Molarity = 1.9M

Mole of solute =.?

Molarity = mole /Volume

1.9 = mole / 1.4

Cross multiply

Mole = 1.9 x 1.4

Mole = 2.66 moles

Therefore, the mole of the solute present in the solution is 2.66 moles.

16. Data obtained from the question include the following:

Mole of solute = 0.46 mole

Molarity = 0.22M

Volume of solvent (water) =.?

Molarity = mole /Volume

0.22 = 0.46/Volume

Cross multiply

0.22 x Volume = 0.46

Divide both side 0.22

Volume = 0.46/0.22

Volume = 2.09L

Therefore, 2.09L of water is required.

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3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

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Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

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