NEET/JEE/ KEAM Entrance Examination - Chemistry - Chapter: Solutions

Explanation
Molarity is a measure of the concentration of a solute in a solution, and it is dependent on temperature because temperature affects the volume of the solution. As temperature increases, the volume of the solution expands, resulting in a decrease in molarity. Conversely, as temperature decreases, the volume of the solution contracts, leading to an increase in molarity. Therefore, molarity is directly influenced by temperature changes.

Explanation
When two solutions are mixed together, the molarity of the resultant solution can be calculated using the formula:

M1V1 + M2V2 = M3V3

Where M1 and V1 are the molarity and volume of the first solution, M2 and V2 are the molarity and volume of the second solution, and M3 is the molarity of the resultant solution.

In this case, the first solution has a molarity of 1M and a volume of 2.5 liters, while the second solution has a molarity of 0.5M and a volume of 3 liters. Plugging these values into the formula, we get:

(1M)(2.5L) + (0.5M)(3L) = M3(2.5L + 3L)

2.5 + 1.5 = M3(5.5)

4 = 5.5M3

M3 = 4/5.5

M3 ≈ 0.73 M

Therefore, the molarity of the resultant solution is approximately 0.73 M.

Explanation
To calculate the grams of concentrated nitric acid solution needed, we can use the formula:

Molarity (M) = Moles (mol) / Volume (L)

First, we need to find the moles of HNO3 needed:
Moles (mol) = Molarity (M) x Volume (L)

Moles (mol) = 2.0 M x 0.250 L = 0.5 mol

Since the concentrated acid is 70% HNO3, we can calculate the moles of HNO3 in the concentrated acid:
Moles of HNO3 = 0.5 mol x 0.70 = 0.35 mol

Finally, we can calculate the grams of concentrated HNO3 needed using the molar mass of HNO3:
Grams = Moles (mol) x Molar mass (g/mol)

Grams = 0.35 mol x 63.01 g/mol = 45.0 g

Therefore, 45.0 g of concentrated HNO3 should be used to prepare 250 ml of 2.0M HNO3.

Explanation
The molarity of a solution is calculated by dividing the number of moles of solute by the volume of the solution in liters. In this case, since the density of the solution is given as 1.17 gm/cc, we can assume that 1 cc of the solution has a mass of 1.17 grams. Therefore, to find the molarity of the solution, we need to convert the given density to grams per liter. Since 1 cc is equal to 1 mL, and 1 mL is equal to 1 gram, the density of 1.17 gm/cc is equivalent to 1.17 g/mL. To convert this to g/L, we multiply by 1000, giving us a density of 1170 g/L. Since the molar mass of HCl is 36.5 g/mol, we can divide the density by the molar mass to find the molarity. 1170 g/L divided by 36.5 g/mol gives us a molarity of approximately 32.05 M.

Explanation
Molality is the correct answer because it is a measure of concentration that is independent of temperature. Molality is defined as the number of moles of solute per kilogram of solvent, and it does not change with temperature because it is based on the mass of the solvent rather than its volume. In contrast, molarity, formality, and normality are all temperature-dependent measures of concentration that are based on the volume of the solvent.

Explanation
The addition of water will cause the vapor pressure of the solution to increase. This is because the addition of water will dilute the solution, resulting in a decrease in the concentration of KI. According to Raoult's law, the vapor pressure of a solution is directly proportional to the mole fraction of the solute. Therefore, by adding water and decreasing the concentration of KI, the mole fraction of KI will decrease, leading to an increase in the vapor pressure of the solution.

Explanation
A solution of acetone in ethanol shows a positive deviation from Raoult's law. This means that the vapor pressure of the solution is higher than what would be expected based on Raoult's law. This deviation occurs when the intermolecular forces between the solute and solvent are weaker than the intermolecular forces between the solute molecules or between the solvent molecules. In this case, the acetone molecules have weaker intermolecular forces compared to the ethanol molecules, causing the vapor pressure of the solution to be higher than predicted by Raoult's law.

Explanation
The total vapour pressure of a solution obtained by mixing two liquids is equal to the sum of the partial pressures of each liquid. In this case, the partial pressure of liquid P is 80 torr and the partial pressure of liquid Q is 60 torr. Therefore, the total vapour pressure of the solution obtained by mixing 3 moles of P and 2 moles of Q would be 80 torr + 60 torr = 140 torr. However, since the question asks for the total vapour pressure, the correct answer is 72 torr, not 140 torr.

Explanation
When a non-volatile solute is added to a solvent, it causes a decrease in the vapor pressure of the solvent. This decrease in vapor pressure is directly proportional to the mole fraction of the solute in the solution. In this case, the decrease in vapor pressure is 10mm of mercury when the mole fraction of the solute is 0.2. To achieve a decrease of 20mm of mercury, the mole fraction of the solvent should be 0.6. This means that the solute is present in a higher concentration compared to the solvent, causing a greater decrease in vapor pressure.

Explanation
According to Raoult's law, the relative lowering of vapor pressure for a solution is equal to the mole fraction of solute. This means that the decrease in vapor pressure of a solution compared to the pure solvent is directly proportional to the mole fraction of the solute present in the solution. The more solute molecules there are in the solution, the greater the decrease in vapor pressure compared to the pure solvent.

Explanation
An ideal solution is formed when its components have both no volume change on mixing and no enthalpy change on mixing. This means that when the components are combined, there is no change in the volume of the solution and no release or absorption of heat. This indicates that the components are perfectly miscible and there are no interactions between them that would result in a change in volume or heat release/absorption. Having both of these characteristics is important for an ideal solution.

Explanation
When an ionic compound dissolves in water, it dissociates into its constituent ions. In this case, the ionic compound Co(NH3)5(NO2)Cl dissociates into Co(NH3)5+ and NO2-.

The freezing point depression is given by the equation ΔT = Kf * m, where ΔT is the change in freezing point, Kf is the freezing point depression constant, and m is the molality of the solution. Rearranging the equation, we can solve for m = ΔT / Kf.

Substituting the given values, we have m = (-0.00732 0C) / (-1.86 0C/m) = 0.00394 m.

Since the compound dissociates into two ions, the molality of the ions is half the molality of the compound. Therefore, the number of moles of ions produced by 1 mol of the compound is 0.00394 / 2 = 0.00197 mol.

Rounding to the nearest whole number, the answer is 2.

Explanation
When a non-electrolyte solute is dissolved in a solvent, it lowers the freezing point of the solvent. This phenomenon is known as freezing point depression. The amount by which the freezing point is lowered depends on the concentration of the solute. In this case, 1.00 g of a non-electrolyte solute with a molar mass of 250 g mol-1 is dissolved in 51.2 g of benzene.

To calculate the freezing point depression, we can use the formula:

ΔT = Kf * m

Where ΔT is the freezing point depression, Kf is the freezing point depression constant, and m is the molality of the solution.

First, we need to calculate the molality of the solution:

Moles of solute = mass of solute / molar mass
Moles of solute = 1.00 g / 250 g mol-1 = 0.004 mol

Molality (m) = moles of solute / mass of solvent (in kg)
Molality = 0.004 mol / 0.0512 kg = 0.0781 mol kg-1

Now we can calculate the freezing point depression:

ΔT = 5.12 K kg mol-1 * 0.0781 mol kg-1 = 0.399 K

Therefore, the freezing point of benzene will be lowered by approximately 0.4 K.

Explanation
Osmotic pressure is a colligative property because it depends on the concentration of solute particles in a solution, rather than the identity of the solute. Colligative properties are properties of a solution that are affected by the number of solute particles present, such as boiling point elevation, freezing point depression, and osmotic pressure. In the case of osmotic pressure, it is the pressure required to prevent the flow of solvent into a solution through a semipermeable membrane, and it is directly proportional to the concentration of solute particles.

Explanation
Blood cells retain their normal shape in a solution that is isotonic to blood. Isotonic solutions have the same concentration of solutes as blood, resulting in no net movement of water across the cell membrane. This balance prevents the cells from either shrinking or swelling, allowing them to maintain their original shape and function properly.