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Dalton's Law of Partial Pressures

The partial p...

Question

The partial pressure of hydrogen in a flask containing $2 g$ of hydrogen and $32 g$ of sulphur dioxide is:

1 / 16^{t h}

of the total pressure

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1 / 9^{t h}

of the total pressure

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2 / 3^{r d}

of the total pressure

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1 / 8^{t h}

of the total pressure

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Solution

The correct option is C $2 / 3^{r d}$ of the total pressure
$n_{H_{2}} = \frac{2 g}{2 g/mol} = 1 mol$
and $n_{S O_{2}} = \frac{32 g}{64 g/mol} = 0.5 mol$
So, partial pressure willl be:
$∴ P_{H_{2}} = \frac{n_{H_{2}}}{(n_{H_{2}} + n_{S O_{2}})} \times P_{T} = \frac{1}{(1 + 0.5)} \times P_{T} = \frac{2}{3} P_{T}$
Therefore, the partial pressure of hydrogen is ${2 / 3}^{r d}$ of total pressure.

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Similar questions

Q. The partial pressure of hydrogen in a flask containing

2 g

of hydrogen and

32 g

of sulphur dioxide is:

Q. The partial pressure of hydrogen in a flask containing two grams of hydrogen and 32 gm of sulphur dioxide is :

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Q. The partial pressure of hydrogen in a flask containing

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(a) A vessel contains N molecules of oxygen at a certain temperature and pressure. How many molecules of sulphur dioxide can the vessel accommodate at the same temperature and pressure?
(b) Each of the two flasks contain 2.0 g of gas at the same temperature and pressure. One flask contains oxygen and the other hydrogen.
(i) Which sample contains the greater number of molecules?
(ii) If the hydrogen sample contains N molecules, how many molecules are present in oxygen sample?

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Dalton's Law of Partial Pressures

Standard XI Chemistry

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The partial pressure of hydrogen in a flask containing 2 g of hydrogen and 32 g of sulphur dioxide is:

The correct option is C 2/3rd of the total pressurenH2=2 g2 g/mol=1 mol and nSO2=32 g64 g/mol=0.5 mol So, partial pressure willl be: ∴ PH2=nH2(nH2+nSO2)×PT=1(1+0.5)×PT=23PT Therefore, the partial pressure of hydrogen is 2/3rd of total pressure.

The partial pressure of hydrogen in a flask containing $2 g$ of hydrogen and $32 g$ of sulphur dioxide is: