Dalton's Law of Partial Pressure

A gaseous mixture was prepared by taking equal mole of  $COand{N}_2.$ If the total pressure of the mixture was found 1 atmosphere, the partial pressure of the nitrogen  $(N_2)$ in the mixture is:

If 500ml of gas A at 400 torrs and 666.6 ml of B at 600 torrs are placed in a 3 liter flask, the pressure of the system will be

If 500ml of gas A at 400 torr and 666.6 ml of B at 600 torr are placed in a 3 litre flask, the pressure of the system will be

A gaseous mixture of $2$ moles of $\text{A}$, $3$ moles of $\text{B}$, $5$ moles of $\text{C}$ and $10$ moles of $\text{D}$ is contained in a vessel. Assuming that gases are ideal and the partial pressure of $\text{C}$ is $1.5$ atm, total pressure is

If ${\mathrm{X}}_{\mathrm{M}}, {\mathrm{X}}_{\mathrm{P}}$ and ${\mathrm{X}}_{\mathrm{V}}$ are mole fraction, pressure fraction and volume fraction, respectively, of a gaseous mixture, then

$16 \mathrm{g}$ each of ${\mathrm{H}}_2$, $\mathrm{He}$ and ${\mathrm{O}}_2$ are present in a container exerting $10 \mathrm{atm}$ pressure at $\mathrm{T}\left(\mathrm{K}\right)$. The pressure in atm exerted by$16 \mathrm{g}$ each of $\mathrm{He}$ and ${\mathrm{O}}_2$ in the second container of same volume and temperature is

A mixture of nitrogen and water vapour is admitted to a flask that contains a solid drying agent. Immediately after admission, the pressure in the flask is $760 \mathrm{Torr}$. After standing for some hours, the pressure reaches a steady value of $745 \mathrm{Torr}$. Calculate the mole percent of water vapour in the original mixture.

A mixture of ${\mathrm{N}}_2$ and $\mathrm{Ar}$ gases in a cylinder contains $7\mathrm{g}$ of ${\mathrm{N}}_2$ and $8\mathrm{g}$ of $\mathrm{Ar}$. If the total pressure of the mixture of the gases in the cylinder is $27$ bar, the partial pressure of ${\mathrm{N}}_2$ is:
[Use atomic masses (in $\mathrm{g} {\mathrm{mol}}^{–1}$): $\mathrm{N}=14,\mathrm{Ar}=40$]

Given $\mathrm{p}_{\mathrm{CC}_{4}}^{0}=100$ torr, $\mathrm{p}_{\mathrm{H}_{2} \mathrm{O}}^{0}=300$ torr
What will be total pressure (in torr) for immiscible mixture after opening the valve? Give your answer after
dividing by $20$.

60 g of gaseous ${\mathrm{C}}_2{\mathrm{H}}_6$ is mixed with 28 g of carbon monoxide. The pressure of the resulting gaseous mixture is 3 atm. The partial pressure in atm of ${\mathrm{C}}_2{\mathrm{H}}_6$ in the mixture is

If the ratio of masses of $\mathrm{S}{\mathrm{O}}_3$ and ${\mathrm{O}}_2$ gases confined in a vessel is $1:1$, then the ratio of their partial pressure would be

To which of the followings the Dalton's law of partial pressures is not applicable?

Given a mixture with $0.5$ mole of gas A and $x$ moles of gas B. Total pressure is $200Pa$ at $1000 K$ temperature in a vessel of volume $10m^3$ . Then, find $x.$ (R universal gas constant)

The density of a mixture of $O_2$ and $N_2$ is $1.3 g{L}^{-1}$ at STP. Calculate partial pressure of $O_2.$

Gaseous benzene reacts with hydrogen gas in presence of a nickel catalyst to form gaseous cyclohexane according to the reaction,

$C_6{H}_6\left(g\right)+3H_2\left(g\right)\to C_6{H}_{12}\left(g\right)$

A mixture of  $C_6{H}_6$ and excess $H_2$ has a pressure of 60 mm of Hg in an unknown volume. After the gas had been passed over a nickel catalyst and all the benzene converted to cyclohexane, the pressure of the gas was 30 mm of Hg in the same volume at the same temperature. The fraction of $C_6{H}_6$ (by volume) present in the original volume is

A sample of air contains only ${\mathrm{N}}_2,\mathrm{}{\mathrm{O}}_2$ and ${\mathrm{H}}_2\mathrm{O}$. It is saturated with water vapours and the total pressure is 640 torr. The vapour pressure of water is 40 torr and the molar ratio of ${\mathrm{N}}_2\mathrm{}:{\mathrm{O}}_2$ is 3 : 1. The partial pressure of ${\mathrm{N}}_2$ in the sample is

Statement-1: In an empty container, equal weights of methane and hydrogen are mixed at ${25}^{\mathrm{o}}\mathrm{C}$ . The fraction of the total pressure exerted by methane is $\frac{1}{5}$

Statement-2: ${\mathrm{P}}_{{\mathrm{CH}}_4}=\text{Mole fraction of}\mathrm{ }{\mathrm{CH}}_4\times {\mathrm{P}}_{\mathrm{total}}$

Statement = Considering van der Waals' equation of state for a real gas $\left(P+n^2 \frac{a}{V^2}\right) \left(V-nb\right)=nRT;$ the constant 'a' for $O_2$ is less than that for $H_2{O}_{\left(g\right)}$

Explanation = The molar mass of $O_2$ is almost twice that of $H_{2}O$

$2 \mathrm{g}$ of gas $\mathrm{A}$ is introduced to an evacuated flask at $25°\mathrm{C}$. The pressure of the gas is $1 \mathrm{atm}$. Now $3 \mathrm{g}$ of another gas $\mathrm{B}$ is introduced in the same flask. The total pressure becomes $1.5 \mathrm{atm}$. Calculate the volume of the vessel if $\mathrm{A}$ is ${\mathrm{O}}_2$.

A flask of capacity one litre contains $N{H}_3$ at $1 \mathrm{atm}$ & $25 °\mathrm{C}$ . A spark is passed through until all the $N{H}_3$ is decomposed into $N_2$ & $H_2$ . Calculate the pressure of gases left at $25 °\mathrm{C}$ .