276970
The density of oxygen gas at $5 \mathrm{~atm}$ and $127^{\circ} \mathrm{C}$ will be
1 $2.80 \mathrm{~g} / \mathrm{L}$
2 $4.88 \mathrm{~g} / \mathrm{L}$
3 $1.49 \mathrm{~g} / \mathrm{L}$
4 $5.60 \mathrm{~g} / \mathrm{L}$
Explanation:
Density of a gas, $\mathrm{d}=\frac{\mathrm{pM}}{\mathrm{RT}}$ $\because \mathrm{p}=$ pressure $=5 \mathrm{~atm}$ $\begin{aligned} & \mathrm{M}=\text { molar mass of } \mathrm{O}_{2} \text { gas }=32 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{R}=0.082 \mathrm{~L} \text { atm } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~T}=400 \mathrm{~K} \\ & \mathrm{~d}=\frac{5 \times 32}{0.082 \times(273+127)} \\ & =4.878 \approx 4.88 \mathrm{gL}^{-1} \end{aligned}$
AP EAMCET (Engg.) 17.09.2020 Shift-II
SOLUTIONS
276972
The density of $2 \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \mathrm{~g} / \mathrm{cm}^{3}$. The molality of the solution is [Given that molecular mass of $\mathrm{NaOH}=40 \mathrm{~g}$ mol $^{-1}$ ]
276973
In water saturated air, the mole fraction of water vapour is 0.02 , If the total pressure of the saturated air is $1.2 \mathrm{~atm}$, the partial pressure of dry air is
1 $1.18 \mathrm{~atm}$
2 $1.76 \mathrm{~atm}$
3 $1.176 \mathrm{~atm}$
4 $0.98 \mathrm{~atm}$
Explanation:
Given, mole fraction of water vapour $\left(\mathrm{X}_{\mathrm{H}_{2} \mathrm{O}}\right)=0.02$ Total pressure of saturated air $\left(\mathrm{P}_{\mathrm{T}}\right)=1.2 \mathrm{~atm}$ $\mathrm{P}_{\mathrm{O}_{2}}=\mathrm{P}_{\mathrm{T}} \times \mathrm{H}_{2} \mathrm{O}$ So, $\begin{aligned} \mathrm{Po}_{2} & =1.2 \times 0.02 \\ & =0.024 \end{aligned}$ Then partial pressure of dry air $=1.2-0.024$ \begin{array}{r} \mathrm{X}=1.176 \mathrm{~atm} \end{array}
(Odisha NEET-2019)
SOLUTIONS
276975
Calculate molarity of one litre solution of 22.2 gm $\mathrm{CaCl}_{2}$.
1 $0.4 \mathrm{M}$
2 $0.2 \mathrm{M}$
3 $0.8 \mathrm{M}$
4 $0.6 \mathrm{M}$
Explanation:
Applying molarity formula. Molarity $=\frac{\text { Number of moles solute }}{\text { Volume of solution }(\mathrm{L})}$ $\begin{aligned} & =\frac{22.2}{111 \times 1} \quad\left(\therefore \mathrm{CaCl}_{2} \text { molar mass }=111 \mathrm{~g}\right) \\ & =0.2 \mathrm{M} \end{aligned}$
276970
The density of oxygen gas at $5 \mathrm{~atm}$ and $127^{\circ} \mathrm{C}$ will be
1 $2.80 \mathrm{~g} / \mathrm{L}$
2 $4.88 \mathrm{~g} / \mathrm{L}$
3 $1.49 \mathrm{~g} / \mathrm{L}$
4 $5.60 \mathrm{~g} / \mathrm{L}$
Explanation:
Density of a gas, $\mathrm{d}=\frac{\mathrm{pM}}{\mathrm{RT}}$ $\because \mathrm{p}=$ pressure $=5 \mathrm{~atm}$ $\begin{aligned} & \mathrm{M}=\text { molar mass of } \mathrm{O}_{2} \text { gas }=32 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{R}=0.082 \mathrm{~L} \text { atm } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~T}=400 \mathrm{~K} \\ & \mathrm{~d}=\frac{5 \times 32}{0.082 \times(273+127)} \\ & =4.878 \approx 4.88 \mathrm{gL}^{-1} \end{aligned}$
AP EAMCET (Engg.) 17.09.2020 Shift-II
SOLUTIONS
276972
The density of $2 \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \mathrm{~g} / \mathrm{cm}^{3}$. The molality of the solution is [Given that molecular mass of $\mathrm{NaOH}=40 \mathrm{~g}$ mol $^{-1}$ ]
276973
In water saturated air, the mole fraction of water vapour is 0.02 , If the total pressure of the saturated air is $1.2 \mathrm{~atm}$, the partial pressure of dry air is
1 $1.18 \mathrm{~atm}$
2 $1.76 \mathrm{~atm}$
3 $1.176 \mathrm{~atm}$
4 $0.98 \mathrm{~atm}$
Explanation:
Given, mole fraction of water vapour $\left(\mathrm{X}_{\mathrm{H}_{2} \mathrm{O}}\right)=0.02$ Total pressure of saturated air $\left(\mathrm{P}_{\mathrm{T}}\right)=1.2 \mathrm{~atm}$ $\mathrm{P}_{\mathrm{O}_{2}}=\mathrm{P}_{\mathrm{T}} \times \mathrm{H}_{2} \mathrm{O}$ So, $\begin{aligned} \mathrm{Po}_{2} & =1.2 \times 0.02 \\ & =0.024 \end{aligned}$ Then partial pressure of dry air $=1.2-0.024$ \begin{array}{r} \mathrm{X}=1.176 \mathrm{~atm} \end{array}
(Odisha NEET-2019)
SOLUTIONS
276975
Calculate molarity of one litre solution of 22.2 gm $\mathrm{CaCl}_{2}$.
1 $0.4 \mathrm{M}$
2 $0.2 \mathrm{M}$
3 $0.8 \mathrm{M}$
4 $0.6 \mathrm{M}$
Explanation:
Applying molarity formula. Molarity $=\frac{\text { Number of moles solute }}{\text { Volume of solution }(\mathrm{L})}$ $\begin{aligned} & =\frac{22.2}{111 \times 1} \quad\left(\therefore \mathrm{CaCl}_{2} \text { molar mass }=111 \mathrm{~g}\right) \\ & =0.2 \mathrm{M} \end{aligned}$
276970
The density of oxygen gas at $5 \mathrm{~atm}$ and $127^{\circ} \mathrm{C}$ will be
1 $2.80 \mathrm{~g} / \mathrm{L}$
2 $4.88 \mathrm{~g} / \mathrm{L}$
3 $1.49 \mathrm{~g} / \mathrm{L}$
4 $5.60 \mathrm{~g} / \mathrm{L}$
Explanation:
Density of a gas, $\mathrm{d}=\frac{\mathrm{pM}}{\mathrm{RT}}$ $\because \mathrm{p}=$ pressure $=5 \mathrm{~atm}$ $\begin{aligned} & \mathrm{M}=\text { molar mass of } \mathrm{O}_{2} \text { gas }=32 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{R}=0.082 \mathrm{~L} \text { atm } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~T}=400 \mathrm{~K} \\ & \mathrm{~d}=\frac{5 \times 32}{0.082 \times(273+127)} \\ & =4.878 \approx 4.88 \mathrm{gL}^{-1} \end{aligned}$
AP EAMCET (Engg.) 17.09.2020 Shift-II
SOLUTIONS
276972
The density of $2 \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \mathrm{~g} / \mathrm{cm}^{3}$. The molality of the solution is [Given that molecular mass of $\mathrm{NaOH}=40 \mathrm{~g}$ mol $^{-1}$ ]
276973
In water saturated air, the mole fraction of water vapour is 0.02 , If the total pressure of the saturated air is $1.2 \mathrm{~atm}$, the partial pressure of dry air is
1 $1.18 \mathrm{~atm}$
2 $1.76 \mathrm{~atm}$
3 $1.176 \mathrm{~atm}$
4 $0.98 \mathrm{~atm}$
Explanation:
Given, mole fraction of water vapour $\left(\mathrm{X}_{\mathrm{H}_{2} \mathrm{O}}\right)=0.02$ Total pressure of saturated air $\left(\mathrm{P}_{\mathrm{T}}\right)=1.2 \mathrm{~atm}$ $\mathrm{P}_{\mathrm{O}_{2}}=\mathrm{P}_{\mathrm{T}} \times \mathrm{H}_{2} \mathrm{O}$ So, $\begin{aligned} \mathrm{Po}_{2} & =1.2 \times 0.02 \\ & =0.024 \end{aligned}$ Then partial pressure of dry air $=1.2-0.024$ \begin{array}{r} \mathrm{X}=1.176 \mathrm{~atm} \end{array}
(Odisha NEET-2019)
SOLUTIONS
276975
Calculate molarity of one litre solution of 22.2 gm $\mathrm{CaCl}_{2}$.
1 $0.4 \mathrm{M}$
2 $0.2 \mathrm{M}$
3 $0.8 \mathrm{M}$
4 $0.6 \mathrm{M}$
Explanation:
Applying molarity formula. Molarity $=\frac{\text { Number of moles solute }}{\text { Volume of solution }(\mathrm{L})}$ $\begin{aligned} & =\frac{22.2}{111 \times 1} \quad\left(\therefore \mathrm{CaCl}_{2} \text { molar mass }=111 \mathrm{~g}\right) \\ & =0.2 \mathrm{M} \end{aligned}$
276970
The density of oxygen gas at $5 \mathrm{~atm}$ and $127^{\circ} \mathrm{C}$ will be
1 $2.80 \mathrm{~g} / \mathrm{L}$
2 $4.88 \mathrm{~g} / \mathrm{L}$
3 $1.49 \mathrm{~g} / \mathrm{L}$
4 $5.60 \mathrm{~g} / \mathrm{L}$
Explanation:
Density of a gas, $\mathrm{d}=\frac{\mathrm{pM}}{\mathrm{RT}}$ $\because \mathrm{p}=$ pressure $=5 \mathrm{~atm}$ $\begin{aligned} & \mathrm{M}=\text { molar mass of } \mathrm{O}_{2} \text { gas }=32 \mathrm{~g} \mathrm{~mol}^{-1} \\ & \mathrm{R}=0.082 \mathrm{~L} \text { atm } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{~T}=400 \mathrm{~K} \\ & \mathrm{~d}=\frac{5 \times 32}{0.082 \times(273+127)} \\ & =4.878 \approx 4.88 \mathrm{gL}^{-1} \end{aligned}$
AP EAMCET (Engg.) 17.09.2020 Shift-II
SOLUTIONS
276972
The density of $2 \mathrm{M}$ aqueous solution of $\mathrm{NaOH}$ is $1.28 \mathrm{~g} / \mathrm{cm}^{3}$. The molality of the solution is [Given that molecular mass of $\mathrm{NaOH}=40 \mathrm{~g}$ mol $^{-1}$ ]
276973
In water saturated air, the mole fraction of water vapour is 0.02 , If the total pressure of the saturated air is $1.2 \mathrm{~atm}$, the partial pressure of dry air is
1 $1.18 \mathrm{~atm}$
2 $1.76 \mathrm{~atm}$
3 $1.176 \mathrm{~atm}$
4 $0.98 \mathrm{~atm}$
Explanation:
Given, mole fraction of water vapour $\left(\mathrm{X}_{\mathrm{H}_{2} \mathrm{O}}\right)=0.02$ Total pressure of saturated air $\left(\mathrm{P}_{\mathrm{T}}\right)=1.2 \mathrm{~atm}$ $\mathrm{P}_{\mathrm{O}_{2}}=\mathrm{P}_{\mathrm{T}} \times \mathrm{H}_{2} \mathrm{O}$ So, $\begin{aligned} \mathrm{Po}_{2} & =1.2 \times 0.02 \\ & =0.024 \end{aligned}$ Then partial pressure of dry air $=1.2-0.024$ \begin{array}{r} \mathrm{X}=1.176 \mathrm{~atm} \end{array}
(Odisha NEET-2019)
SOLUTIONS
276975
Calculate molarity of one litre solution of 22.2 gm $\mathrm{CaCl}_{2}$.
1 $0.4 \mathrm{M}$
2 $0.2 \mathrm{M}$
3 $0.8 \mathrm{M}$
4 $0.6 \mathrm{M}$
Explanation:
Applying molarity formula. Molarity $=\frac{\text { Number of moles solute }}{\text { Volume of solution }(\mathrm{L})}$ $\begin{aligned} & =\frac{22.2}{111 \times 1} \quad\left(\therefore \mathrm{CaCl}_{2} \text { molar mass }=111 \mathrm{~g}\right) \\ & =0.2 \mathrm{M} \end{aligned}$
