If the density of lake water is 1.25 g mL–1 and contains 92 g of Na+ ions per kg of water, then the molality of Na+ ions will be:
| 1. | 3.24 molal | 2. | 4 molal |
| 3. | 5 molal | 4. | 3.5 molal |
The partial pressure of ethane over a solution containing 6.56 × 10–2 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, the partial pressure of the gas will be:
1. 0.76 bar
2. 0.16 bar
3. 1.16 bar
4. 3.16 bar
The mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN will be:
1. 1.424%
2. 4.424%
3. 5.124%
4. 2.124%
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphene generally given is 1.5 mg.
Calculate the mass of 1.5 × 10−3m aqueous solution required for the above dose.
1. 13.22 g
2. 3.22 g
3. 11.22 g
4. 9.22 g
The amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol is:
| 1. | 4.57 g | 2. | 3.57 g |
| 3. | 1.57 g | 4. | 12.57 g |
Henry’s law constant for the solution of methane in benzene at 298 K is 4.27 × 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg will be:
1. 1.85 × 10–5
2. 192 × 10–4
3. 178 × 10–5
4. 18.7 × 10–5
Air consists primarily of nitrogen, which occupies approximately 79% by volume at 298 K under normal conditions. Water is in equilibrium with air at a total pressure of 10 atm and a temperature 298 K. The Henry's law constant for nitrogen in water at 298 K is given as 6.51 × 10⁷ mm Hg.
The mole fraction of dissolved nitrogen in the water under these conditions is:
| 1. | 12.4 × 10−5
|
2. | 9.22 × 10−5
|
| 3. | 3.54 × 10−5
|
4. | 4.96 × 10−5 |
To minimize the painful effects accompanying deep sea diving, oxygen diluted with less soluble helium gas is used as breathing gas by the divers. This is an example of the application of:
| 1. | Raoult's law | 2. | Henry's law |
| 3. | Ideal gas Equation | 4. | All of the above |
| 1. | \(\text X_{\text{mole fraction}}=\frac{\text n_{\text{solute}}}{\text n_{\text{solution}}}\) |
| 2. | \(\text{Molarity}=\frac{\text{amount of solute (g)}}{\text{volume of solution (mL)}}\) |
| 3. | \(\text{Molality}=\frac{\text{Number of mole of solute}}{\text{amount of solvent (kg)}}\) |
| 4. | \(\text{Mass percentage}=\frac{\text{mass of the component in the solution}}{\text{Total mass of the solution}}\times100 \) |
The density of 68 % nitric acid by mass in an aqueous solution is 1.504 g mL–1. The molarity of the acid solution would be:
1. 15.24 M
2. 16.23 M
3. 14.52 M
4. 13.45 M