NEET Test Series from KOTA - 10 Papers In MS WORD
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Some Basic Concepts of Chemistry
228887
The mass of oxygen gas which occupies 5.6 litres at STP would be
1 the gram atomic mass of oxygen
2 one-fourth of the gram atomic mass of oxygen
3 double the gram atomic mass of oxyen
4 half of the gram atomic mass of oxygen.
Explanation:
32 g of oxygen at STP occupies $=22.4 \mathrm{~L}$ $5.6 \mathrm{~L}$ of oxygen at STP occupies $\begin{aligned} & =\frac{32}{22.4} \times 5.6=8 \text { g of oxygen } \\ & =\frac{1}{2} \times \text { Gramatomic mass of oxygen } \end{aligned}$
COMEDK-2015
Some Basic Concepts of Chemistry
228889
A gas ' $X$ ' is dissolved in water at 2 bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is :
1 0.04
2 0.98
3 0.96
4 0.02
Explanation:
Given that, $\mathrm{P}_1=2$ bar $\mathrm{P}_2=2 \times 2=4 \text { bar, } \mathrm{X}_1=0.02$ $\because$ We know that $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{X}_1}{\mathrm{X}_2}$ Putting the value, we get- $\begin{aligned} \frac{2}{4} & =\frac{0.02}{\mathrm{X}_2} \\ \mathrm{X}_2 & =0.04 \end{aligned}$ Therefore, In solution, $\mathrm{X}_{\text {solute }}+\mathrm{X}_{\text {solvent }}=1$ $0.04+\mathrm{X}_{\text {solvent }}=1$ $\text { or } \mathrm{X}_{\text {solvent }}=1-0.04=0.96$
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228890
A metal oxide has the empirical formula, $\mathbf{M}_{0.96} \mathbf{O}_{1.00}$. What will be the percentage of $\mathrm{M}^{2+}$ ions in the crystal?
1 90.67
2 91.67
3 8.33
4 9.33
Explanation:
$\mathrm{M}_{0.96} \mathrm{O}_{1.00}$ may contains $\mathrm{M}^{2+}, \mathrm{M}^{3+}$ ions. Amount of $\mathrm{M}^{2+}$ ions in crystals $=\mathrm{x}$ Amount of $\mathrm{M}^{3+}$ ions will be $=0.96-\mathrm{x}$ We know that negative and positive ions are equal. $\begin{aligned} & \therefore \quad 2 \mathrm{x}+3(0.96-\mathrm{x})=2 \\ & 2 \mathrm{x}+3(0.96-\mathrm{x})-2=0 \\ & 2 \mathrm{x}+2.88-3 \mathrm{x}-2=0 \\ & 0.88-\mathrm{x}=0 \\ & \Rightarrow \quad \mathrm{x}=0.88 \\ & \% \text { of } \mathrm{M}^{2+} \text { ions in the crystal }=\frac{0.88}{0.96} \times 100=91.67 \% \\ & \end{aligned}$
AMU-2015
Some Basic Concepts of Chemistry
228892
20.0 $g$ of magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 \mathrm{~g}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. of $\mathrm{Mg}=24$ )
1 96
2 60
3 84
4 75
Explanation:
$\mathrm{MgCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2$ (g) Molar mass of $\mathrm{MgCO}_3=24+12+48=84 \mathrm{~g} / \mathrm{mol}$ Molar mass of $\mathrm{MgO}=24+16=40 \mathrm{~g} / \mathrm{mol}$ $\therefore 84 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv 40 \mathrm{~g}$ of $\mathrm{MgO}$ And $20 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv \frac{40}{84} \times 20=9.52 \mathrm{~g}$ of $\mathrm{MgO}$ But actual yield $=8 \mathrm{~g}$ of $\mathrm{MgO}$ $\therefore \% \text { purity }=\frac{8}{9.52} \times 100=84 \%$
228887
The mass of oxygen gas which occupies 5.6 litres at STP would be
1 the gram atomic mass of oxygen
2 one-fourth of the gram atomic mass of oxygen
3 double the gram atomic mass of oxyen
4 half of the gram atomic mass of oxygen.
Explanation:
32 g of oxygen at STP occupies $=22.4 \mathrm{~L}$ $5.6 \mathrm{~L}$ of oxygen at STP occupies $\begin{aligned} & =\frac{32}{22.4} \times 5.6=8 \text { g of oxygen } \\ & =\frac{1}{2} \times \text { Gramatomic mass of oxygen } \end{aligned}$
COMEDK-2015
Some Basic Concepts of Chemistry
228889
A gas ' $X$ ' is dissolved in water at 2 bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is :
1 0.04
2 0.98
3 0.96
4 0.02
Explanation:
Given that, $\mathrm{P}_1=2$ bar $\mathrm{P}_2=2 \times 2=4 \text { bar, } \mathrm{X}_1=0.02$ $\because$ We know that $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{X}_1}{\mathrm{X}_2}$ Putting the value, we get- $\begin{aligned} \frac{2}{4} & =\frac{0.02}{\mathrm{X}_2} \\ \mathrm{X}_2 & =0.04 \end{aligned}$ Therefore, In solution, $\mathrm{X}_{\text {solute }}+\mathrm{X}_{\text {solvent }}=1$ $0.04+\mathrm{X}_{\text {solvent }}=1$ $\text { or } \mathrm{X}_{\text {solvent }}=1-0.04=0.96$
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228890
A metal oxide has the empirical formula, $\mathbf{M}_{0.96} \mathbf{O}_{1.00}$. What will be the percentage of $\mathrm{M}^{2+}$ ions in the crystal?
1 90.67
2 91.67
3 8.33
4 9.33
Explanation:
$\mathrm{M}_{0.96} \mathrm{O}_{1.00}$ may contains $\mathrm{M}^{2+}, \mathrm{M}^{3+}$ ions. Amount of $\mathrm{M}^{2+}$ ions in crystals $=\mathrm{x}$ Amount of $\mathrm{M}^{3+}$ ions will be $=0.96-\mathrm{x}$ We know that negative and positive ions are equal. $\begin{aligned} & \therefore \quad 2 \mathrm{x}+3(0.96-\mathrm{x})=2 \\ & 2 \mathrm{x}+3(0.96-\mathrm{x})-2=0 \\ & 2 \mathrm{x}+2.88-3 \mathrm{x}-2=0 \\ & 0.88-\mathrm{x}=0 \\ & \Rightarrow \quad \mathrm{x}=0.88 \\ & \% \text { of } \mathrm{M}^{2+} \text { ions in the crystal }=\frac{0.88}{0.96} \times 100=91.67 \% \\ & \end{aligned}$
AMU-2015
Some Basic Concepts of Chemistry
228892
20.0 $g$ of magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 \mathrm{~g}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. of $\mathrm{Mg}=24$ )
1 96
2 60
3 84
4 75
Explanation:
$\mathrm{MgCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2$ (g) Molar mass of $\mathrm{MgCO}_3=24+12+48=84 \mathrm{~g} / \mathrm{mol}$ Molar mass of $\mathrm{MgO}=24+16=40 \mathrm{~g} / \mathrm{mol}$ $\therefore 84 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv 40 \mathrm{~g}$ of $\mathrm{MgO}$ And $20 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv \frac{40}{84} \times 20=9.52 \mathrm{~g}$ of $\mathrm{MgO}$ But actual yield $=8 \mathrm{~g}$ of $\mathrm{MgO}$ $\therefore \% \text { purity }=\frac{8}{9.52} \times 100=84 \%$
228887
The mass of oxygen gas which occupies 5.6 litres at STP would be
1 the gram atomic mass of oxygen
2 one-fourth of the gram atomic mass of oxygen
3 double the gram atomic mass of oxyen
4 half of the gram atomic mass of oxygen.
Explanation:
32 g of oxygen at STP occupies $=22.4 \mathrm{~L}$ $5.6 \mathrm{~L}$ of oxygen at STP occupies $\begin{aligned} & =\frac{32}{22.4} \times 5.6=8 \text { g of oxygen } \\ & =\frac{1}{2} \times \text { Gramatomic mass of oxygen } \end{aligned}$
COMEDK-2015
Some Basic Concepts of Chemistry
228889
A gas ' $X$ ' is dissolved in water at 2 bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is :
1 0.04
2 0.98
3 0.96
4 0.02
Explanation:
Given that, $\mathrm{P}_1=2$ bar $\mathrm{P}_2=2 \times 2=4 \text { bar, } \mathrm{X}_1=0.02$ $\because$ We know that $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{X}_1}{\mathrm{X}_2}$ Putting the value, we get- $\begin{aligned} \frac{2}{4} & =\frac{0.02}{\mathrm{X}_2} \\ \mathrm{X}_2 & =0.04 \end{aligned}$ Therefore, In solution, $\mathrm{X}_{\text {solute }}+\mathrm{X}_{\text {solvent }}=1$ $0.04+\mathrm{X}_{\text {solvent }}=1$ $\text { or } \mathrm{X}_{\text {solvent }}=1-0.04=0.96$
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228890
A metal oxide has the empirical formula, $\mathbf{M}_{0.96} \mathbf{O}_{1.00}$. What will be the percentage of $\mathrm{M}^{2+}$ ions in the crystal?
1 90.67
2 91.67
3 8.33
4 9.33
Explanation:
$\mathrm{M}_{0.96} \mathrm{O}_{1.00}$ may contains $\mathrm{M}^{2+}, \mathrm{M}^{3+}$ ions. Amount of $\mathrm{M}^{2+}$ ions in crystals $=\mathrm{x}$ Amount of $\mathrm{M}^{3+}$ ions will be $=0.96-\mathrm{x}$ We know that negative and positive ions are equal. $\begin{aligned} & \therefore \quad 2 \mathrm{x}+3(0.96-\mathrm{x})=2 \\ & 2 \mathrm{x}+3(0.96-\mathrm{x})-2=0 \\ & 2 \mathrm{x}+2.88-3 \mathrm{x}-2=0 \\ & 0.88-\mathrm{x}=0 \\ & \Rightarrow \quad \mathrm{x}=0.88 \\ & \% \text { of } \mathrm{M}^{2+} \text { ions in the crystal }=\frac{0.88}{0.96} \times 100=91.67 \% \\ & \end{aligned}$
AMU-2015
Some Basic Concepts of Chemistry
228892
20.0 $g$ of magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 \mathrm{~g}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. of $\mathrm{Mg}=24$ )
1 96
2 60
3 84
4 75
Explanation:
$\mathrm{MgCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2$ (g) Molar mass of $\mathrm{MgCO}_3=24+12+48=84 \mathrm{~g} / \mathrm{mol}$ Molar mass of $\mathrm{MgO}=24+16=40 \mathrm{~g} / \mathrm{mol}$ $\therefore 84 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv 40 \mathrm{~g}$ of $\mathrm{MgO}$ And $20 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv \frac{40}{84} \times 20=9.52 \mathrm{~g}$ of $\mathrm{MgO}$ But actual yield $=8 \mathrm{~g}$ of $\mathrm{MgO}$ $\therefore \% \text { purity }=\frac{8}{9.52} \times 100=84 \%$
228887
The mass of oxygen gas which occupies 5.6 litres at STP would be
1 the gram atomic mass of oxygen
2 one-fourth of the gram atomic mass of oxygen
3 double the gram atomic mass of oxyen
4 half of the gram atomic mass of oxygen.
Explanation:
32 g of oxygen at STP occupies $=22.4 \mathrm{~L}$ $5.6 \mathrm{~L}$ of oxygen at STP occupies $\begin{aligned} & =\frac{32}{22.4} \times 5.6=8 \text { g of oxygen } \\ & =\frac{1}{2} \times \text { Gramatomic mass of oxygen } \end{aligned}$
COMEDK-2015
Some Basic Concepts of Chemistry
228889
A gas ' $X$ ' is dissolved in water at 2 bar pressure. Its mole fraction is 0.02 in solution. The mole fraction of water when the pressure of gas is doubled at the same temperature is :
1 0.04
2 0.98
3 0.96
4 0.02
Explanation:
Given that, $\mathrm{P}_1=2$ bar $\mathrm{P}_2=2 \times 2=4 \text { bar, } \mathrm{X}_1=0.02$ $\because$ We know that $\frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{\mathrm{X}_1}{\mathrm{X}_2}$ Putting the value, we get- $\begin{aligned} \frac{2}{4} & =\frac{0.02}{\mathrm{X}_2} \\ \mathrm{X}_2 & =0.04 \end{aligned}$ Therefore, In solution, $\mathrm{X}_{\text {solute }}+\mathrm{X}_{\text {solvent }}=1$ $0.04+\mathrm{X}_{\text {solvent }}=1$ $\text { or } \mathrm{X}_{\text {solvent }}=1-0.04=0.96$
AP-EAMCET (Engg.)-2015
Some Basic Concepts of Chemistry
228890
A metal oxide has the empirical formula, $\mathbf{M}_{0.96} \mathbf{O}_{1.00}$. What will be the percentage of $\mathrm{M}^{2+}$ ions in the crystal?
1 90.67
2 91.67
3 8.33
4 9.33
Explanation:
$\mathrm{M}_{0.96} \mathrm{O}_{1.00}$ may contains $\mathrm{M}^{2+}, \mathrm{M}^{3+}$ ions. Amount of $\mathrm{M}^{2+}$ ions in crystals $=\mathrm{x}$ Amount of $\mathrm{M}^{3+}$ ions will be $=0.96-\mathrm{x}$ We know that negative and positive ions are equal. $\begin{aligned} & \therefore \quad 2 \mathrm{x}+3(0.96-\mathrm{x})=2 \\ & 2 \mathrm{x}+3(0.96-\mathrm{x})-2=0 \\ & 2 \mathrm{x}+2.88-3 \mathrm{x}-2=0 \\ & 0.88-\mathrm{x}=0 \\ & \Rightarrow \quad \mathrm{x}=0.88 \\ & \% \text { of } \mathrm{M}^{2+} \text { ions in the crystal }=\frac{0.88}{0.96} \times 100=91.67 \% \\ & \end{aligned}$
AMU-2015
Some Basic Concepts of Chemistry
228892
20.0 $g$ of magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0 \mathrm{~g}$ magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. wt. of $\mathrm{Mg}=24$ )
1 96
2 60
3 84
4 75
Explanation:
$\mathrm{MgCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2$ (g) Molar mass of $\mathrm{MgCO}_3=24+12+48=84 \mathrm{~g} / \mathrm{mol}$ Molar mass of $\mathrm{MgO}=24+16=40 \mathrm{~g} / \mathrm{mol}$ $\therefore 84 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv 40 \mathrm{~g}$ of $\mathrm{MgO}$ And $20 \mathrm{~g}$ of $\mathrm{MgCO}_3 \equiv \frac{40}{84} \times 20=9.52 \mathrm{~g}$ of $\mathrm{MgO}$ But actual yield $=8 \mathrm{~g}$ of $\mathrm{MgO}$ $\therefore \% \text { purity }=\frac{8}{9.52} \times 100=84 \%$
