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Alborosie
3 years ago
7

The student collects the H2 (g) produced by the reaction and measures the volume

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

The question is incomplete, the complete question is;

The student collects the H2(g) produced by the reaction and measures its volume over water at 298 K after carefully equalizing the water levels inside and outside the gas-collection tube, as shown in the diagram below. The volume is measured to be 45.6mL . The atmospheric pressure in the lab is measured as 765 torr , and the equilibrium vapor pressure of water at 298 K is 24 torr .(i) The pressure inside the tube due to the H2(g)

Answer:

741 torr

Explanation:

From the question we can see that the atmospheric pressure in the lab is 765 torr.

The vapour pressure of water = 24 torr

From Dalton's law of partial pressure, we know that;

Total pressure = Pressure of the H2 + Partial pressure of water vapour

Therefore;

Pressure of H2 = Total pressure - Partial pressure of water vapour

Pressure of H2 = 765 torr - 24 torr = 741 torr

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A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?
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<u>Answer:</u> The mole fraction of barium chloride in the solution is 0.024

<u>Explanation:</u>

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

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\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g

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<u>Calculating the moles of water:</u>

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol

Mole fraction of a substance is given by:

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

  • <u>For barium chloride:</u>

\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}

\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024

Hence, the mole fraction of barium chloride in the solution is 0.024

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