What is the amount of anhydrous Na2CO3 in 200 ml of 0.25 M solution? (Options: a) 6.225 g b) 66.25 g c) 6.0 g d) 6.625 g)

Molar mass of $Na_2CO_3 = (2 \times 23.0) + 12.0 + (3 \times 16.0) = 46.0 + 12.0 + 48.0 = 106.0$ g/mol.

Given Molarity ($M$) = 0.25 mol/L
Given Volume ($V$) = 200 mL = 0.200 L

Looking at the options, none of them exactly match 5.3 g. Let's recheck the calculation and options.

(a) 6.225 g

(b) 66.25 g

(c) 6.0 g

(d) 6.625 g

There might be a slight discrepancy in the question or options provided. However, if we assume a typo and look for the closest option, 6.0 g (c) is the closest to 5.3 g. Let's re-evaluate if there's any common mistake or alternative interpretation.
If the molarity was 0.25 M and volume 250 mL (0.25 L), then moles = 0.25 * 0.25 = 0.0625 mol. Mass = 0.0625 * 106 = 6.625 g. This matches option (d). It's possible the volume was intended to be 250 mL instead of 200 mL, or the molarity was different. Given the options, 6.625 g is a direct calculation if the volume was 250 mL. Let's assume the question intended for 250 mL or there's a typo in the options. If we strictly follow 200 mL, the answer is 5.3 g. Since 6.625 g is an option, it's highly likely that the volume was meant to be 250 mL.

Let's assume the volume is 250 mL for option (d) to be correct.

Given the options, this is the most plausible intended answer.

The final answer is $\boxed{\text{(d) 6.625 g}}$.

Q.1 (ii) Molarity of pure water is

Density of water is approximately 1 g/mL.
Mass of 1 L (1000 mL) of water = $1000 \text{ mL} \times 1 \text{ g/mL} = 1000 \text{ g}$.

Molar mass of $H_2O = (2 \times 1.0) + 16.0 = 18.0$ g/mol.

Molarity = Moles / Volume (L)

The final answer is $\boxed{\text{(d) 55.5 M}}$.

Q.1 (iii) If 4g of NaOH are dissolved in 500 ml of water, the concentration of the solution will be.

Molar mass of NaOH = $23.0 + 16.0 + 1.0 = 40.0$ g/mol.

Given mass of NaOH = 4 g.

Given volume of solution = 500 mL = 0.500 L.

The final answer is $\boxed{\text{(b) 0.2 M}}$.

Q.1 (iv) Which of the following has the minimum vapour pressure?

Vapour pressure is a colligative property, which means it depends on the number of solute particles in the solution. A higher concentration of solute particles leads to a lower vapour pressure. We need to calculate the effective concentration ($i \times M$) for each option, where $i$ is the van't Hoff factor and $M$ is the molarity.

• (a) 0.1 M $BaCl_2$ solution: $BaCl_2$ dissociates into $Ba^{2+}$ and $2Cl^-$, so $i = 3$. Effective concentration = $3 \times 0.1 \text{ M} = 0.3 \text{ M}$.
• (b) 0.1 M Phenol solution: Phenol ($C_6H_5OH$) is a non-electrolyte (or very weak acid, usually considered non-dissociating for colligative properties). So, $i = 1$. Effective concentration = $1 \times 0.1 \text{ M} = 0.1 \text{ M}$.
• (c) 0.1 M sucrose solution: Sucrose ($C_{12}H_{22}O_{11}$) is a non-electrolyte. So, $i = 1$. Effective concentration = $1 \times 0.1 \text{ M} = 0.1 \text{ M}$.
• (d) 0.1 M solution of sodium chloride ($NaCl$): $NaCl$ dissociates into $Na^+$ and $Cl^-$, so $i = 2$. Effective concentration = $2 \times 0.1 \text{ M} = 0.2 \text{ M}$.

Comparing the effective concentrations:

(a) 0.3 M

(b) 0.1 M

(c) 0.1 M

(d) 0.2 M

The 0.1 M $BaCl_2$ solution has the highest effective concentration (0.3 M), and therefore it will have the minimum vapour pressure.

The final answer is $\boxed{\text{(a) 0.1 M } BaCl_2 \text{ solution}}$.

Q.1 (v) Benzene and toluene form a

Benzene and toluene are structurally similar non-polar organic compounds. When mixed, the intermolecular forces between benzene-benzene, toluene-toluene, and benzene-toluene molecules are very similar. This leads to a solution that closely obeys Raoult's law.

The final answer is $\boxed{\text{(a) Ideal solution}}$.

Q.1 (vi) Which of the following solutions has the highest osmotic pressure?

Osmotic pressure ($\pi$) is a colligative property, directly proportional to the effective molar concentration ($i \times C$), where $i$ is the van't Hoff factor and $C$ is the molar concentration. All options have a concentration of 1 M. We need to find the solution with the highest $i$ value.