A. The molality of a 2.5 g of ethanoic acid (Molar mass : $60 \text{ g mol}^{-1}$) in 75 g of benzene solution is $0.556 \text{ m}$.
B. The molarity of a solution containing 5 g of NaOH (molar mass : $40 \text{ g mol}^{-1}$) in 450 mL of solution is 0.278 M at 298 K.
C. Aquatic species are more comfortable in cold water.
D. The solubility of gas increases with decrease in pressure.
E. For a binary mixture of A and B, the number of moles of A and B are $n_A$ and $n_B$ respectively, the mole fraction of B will be $x_B = \frac{n_A}{n_A + n_B}$.
Choose the correct answer from the options given below :
Statement A Analysis: Molality Calculation
To verify statement A, we calculate the molality of the ethanoic acid solution.
- Mass of solute (ethanoic acid) = 2.5 g
- Molar mass of solute = $60 \text{ g mol}^{-1}$
- Mass of solvent (benzene) = 75 g = 0.075 kg
- Moles of solute = $\frac{\text{Mass of solute}}{\text{Molar mass of solute}} = \frac{2.5 \text{ g}}{60 \text{ g mol}^{-1}} \approx 0.04167 \text{ mol}$
- Molality ($m$) = $\frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.04167 \text{ mol}}{0.075 \text{ kg}} \approx 0.5556 \text{ m}$
The calculated molality is approximately $0.556 \text{ m}$, making statement A correct.
Statement B Analysis: Molarity Calculation
To verify statement B, we calculate the molarity of the NaOH solution.
- Mass of solute (NaOH) = 5 g
- Molar mass of solute = $40 \text{ g mol}^{-1}$
- Volume of solution = 450 mL = 0.450 L
- Moles of solute = $\frac{\text{Mass of solute}}{\text{Molar mass of solute}} = \frac{5 \text{ g}}{40 \text{ g mol}^{-1}} = 0.125 \text{ mol}$
- Molarity ($M$) = $\frac{\text{Moles of solute}}{\text{Volume of solution (L)}} = \frac{0.125 \text{ mol}}{0.450 \text{ L}} \approx 0.2778 \text{ M}$
The calculated molarity is approximately $0.278 \text{ M}$, making statement B correct.
Statement C Analysis: Aquatic Life Comfort
The solubility of gases in water is temperature-dependent. Gases are more soluble in colder water than in warmer water. Aquatic species, like fish, require dissolved oxygen, which is more abundant in cold water. Therefore, aquatic species are more comfortable in cold water.
Statement C is correct.
Statement D Analysis: Gas Solubility vs Pressure
According to Henry's Law, the solubility of a gas in a liquid is directly proportional to the partial pressure of that gas above the liquid. This means solubility increases with increasing pressure.
Statement D, which claims solubility increases with decreasing pressure, is incorrect.
Statement E Analysis: Mole Fraction Formula
For a binary mixture of components A and B, with moles $n_A$ and $n_B$ respectively:
- Total moles = $n_A + n_B$
- The mole fraction of B ($x_B$) is defined as: $x_B = \frac{n_B}{n_A + n_B}$
- The formula provided in the statement, $x_B = \frac{n_A}{n_A + n_B}$, actually represents the mole fraction of A ($x_A$).
Statement E is incorrect.
Conclusion
Based on the analysis, the correct statements are A, B, and C.
Therefore, the correct option is A, B and C only.
Similar Questions
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$K = 4 \times 10^{-10}$
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Important Questions from Physical Chemistry
- Consider the following electrochemical cell :
$\text{Pt} \mid \text{O}_2\text{(g)}(1\text{bar}) \mid \text{HCl(aq)} \parallel \text{M}^{2+}\text{(aq, 1.0 M)} \mid \text{M(s)}$
The pH above which, oxygen gas would start to evolve at anode is _________ (nearest integer).
Given :
$\begin{bmatrix} \text{E}^\circ_{\text{M}^{2+}/\text{M}} = 0.994 \text{ V} \\ \text{E}^\circ_{\text{O}_2/\text{H}_2\text{O}} = 1.23 \text{ V} \end{bmatrix} \text{standard reduction potential}$
and $\frac{\text{RT}}{\text{F}} (2.303) = 0.059 \text{ V}$ at the given condition - X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.
A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - Two liquids A and B form an ideal solution at temperature T K. At T K, the vapour pressures of pure A and B are 55 and $15 \text{ kN m}^{-2}$ respectively. What is the mole fraction of A in solution of A and B in equilibrium with a vapour in which the mole fraction of A is 0.8?
