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4 years ago

Calculate the molarity of the solution to 3 digits.

Chemistry

2 answers:

Lady bird [3.3K]4 years ago

8 0

The correct answer would be 0.254.

Rzqust [24]4 years ago

4 0

<u>Answer:</u> The molarity of the solution is 0.254 mol/L

<u>Explanation:</u>

Molarity is defined as the number of moles of solute present per Liter of solution. It is a unit of concentration. The units used to express molarity is mol/L.

Mathematically,

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times\text{Volume of solution (in L)}}$

We are given:

Given mass of solute = 100 g

Molar mass of solute = 342.28 g/mol

Volume of solution = 1.15 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{100}{342.28\times 1.15}=0.254mol/L$

Hence, the molarity of the solution is 0.254 mol/L

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3 years ago

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3 years ago

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3 years ago

The density of H2O2 is 1.407 g/mL, and the density of O2 is 1.428 g/L. How many liters of O2 can be made from 55 mL H2O2

balandron [24]

Explanation:

mass H2O2 = 55 mL(1.407 g/mL) = 80.85 g

molar mass H2O2 = 2(1.01 g/mol) + 2(16.00 g/mol) = 34.02 g/mol

moles H2O2 = 80.85 g/34.02 g/mol = 2.377 moles H2O2

For each mole of H2O2 you obtain 0.5 mole of O2 (see the equation).

moles O2 = 2.377 moles H2O2 (1 mole O2)/(2 moles H2O2) = 1.188 moles O2

Now, you need the temperature. If you are at STP (273 K, and 1.00 atm) then 1 mole of an ideal gas at STP has a volume of 22.4 L. Without temperature you are not really able to continue. I will assume you are at STP.

Volume O2 = 1.188 moles O2(22.4 L/mole) = 0.0530 L of O2.

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3 years ago

Positive Deviation from Raoult's Law occurs when the vapour pressure of component is greater than what is expected in Raoult's L

8090 [49]

Answer:

All of the above.

Explanation:

In positive deviation from Raoult's Law occur when the vapour pressure of components is greater than what is expected value in Raoult's law.

When a solution is non ideal then it shows positive or negative deviation.

Let two solutions A and B to form non- ideal solutions.let the vapour pressure of component A is $P_A$ and vapour pressure of component B is $P_B$ .

$P^0_A$ = Vapour pressure of component A in pure form

$P^0_B$ = Vapour pressure of component B in pure form

$x_A$ =Mole fraction of component A

$x_B=$ =Mole fraction of component B

The interaction between A- B is less than the interaction A- A and B-B interaction.Therefore, the escaping tendency of liquid molecules in mixture is greater than the escaping tendency in pure form.Hence, the vapour pressure of a mixture is greater than the initial value of vapour pressure.

$P_A >P^0_A\cdot x_A$ , $P_B>P^0_B\cdot x_B$

Therefore, $P_A+P_B >P^0_A\cdot x_A+P^0_B \cdot x_B$

Therefore, the enthalpy of mixing is greater than zero and change in volume is greater than zero.

Hence, option a,b,c and d are true.

6 0

3 years ago