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3 years ago

How many of grams of CuSO4 are in 475ml of a 2.0M aqueous solution

Chemistry

2 answers:

Anettt [7]3 years ago

8 0

Answer:

151.63 g

Explanation:

We first get the number of moles;

Moles = Molarity × volume

= 2.0 × 0.475

= 0.95 moles

1 mole of CuSO4 = 159.609 g/mol

Therefore;

Mass = moles × molar mas

= 0.95 moles × 159.609 g/mol

= 151.63 g

Blizzard [7]3 years ago

3 0

Answer: The mass of copper sulfate present are 151.63 grams

Explanation:

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 2.0 M

Molar mass of copper sulfate = 159.61 g/mol

Volume of solution = 475 mL

Putting values in above equation, we get:

$2.0M=\frac{\text{Mass of copper sulfate}\times 1000}{159.61\times 475}\\\\\text{Mass of copper sulfate}=\frac{2.0\times 159.61\times 475}{1000}=151.63g$

Hence, the mass of copper sulfate present are 151.63 grams

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3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

Answer: The partial pressure of oxygen gas is 2.76 bar

Explanation:

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

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago