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When a certain amount of solid A is dissolved in 100g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is_________.
(Nearest Integer)
When a certain amount of solid A is dissolved in 100g of water at 25°C to make a dilute solution, the vapour pressure of the solution is reduced to one-half of that of pure water. The vapour pressure of pure water is 23.76 mmHg. The number of moles of solute A added is_________.
(Nearest Integer)
1.80g of solute A was dissolved in 62.5cm3 of ethanol and freezing point of the solution was found to be 155.1 K . The molar mass of solute A is __________g mol-1.
[Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm-3. Freezing point depression constant of ethanol is 2.00 K kg mol-1]
1.80g of solute A was dissolved in 62.5cm3 of ethanol and freezing point of the solution was found to be 155.1 K . The molar mass of solute A is __________g mol-1.
[Given: Freezing point of ethanol is 156.0 K. Density of ethanol is 0.80 g cm-3. Freezing point depression constant of ethanol is 2.00 K kg mol-1]
<p><strong>If O<sub>2</sub> gas is bubbled through water at 303 K, the number of millimoles of O<sub>2</sub> gas that dissolve in 1 litre of water is______________. (Nearest integer)</strong></p>
<p><strong>(Given: Henry’s Law constant for O<sub>2</sub> at 303 K is 46.82 k bar and partial pressure of O<sub>2</sub> = 0.920 bar)</strong></p>
<p><strong>(Assume solubility of O<sub>2</sub> in water in too small, nearly negligible)</strong></p>
150g of acetic acid was contaminated with 10.2g ascorbic acid <!-- [if gte mso 9]><xml>
</xml><![endif]--> to lower down its freezing point by (x × 10-1)°C. The value of x is _________. (Nearest integer)
(Given Kf = 3.9 K kg mol-1; molar mass of ascorbic acid = 176 g mol-1]
150g of acetic acid was contaminated with 10.2g ascorbic acid <!-- [if gte mso 9]><xml>
(Given Kf = 3.9 K kg mol-1; molar mass of ascorbic acid = 176 g mol-1]
The osmotic pressure of blood is 7.47 bar at 300K. To inject glucose to a patient intravenously, it has to isotonic with blood. The concentration of glucose solution in gL-1 is __________.
(Molar mass of glucose = 180 g mol-1 R = 0.083 L bar K-1 mol-1) Nearest integer)
The osmotic pressure of blood is 7.47 bar at 300K. To inject glucose to a patient intravenously, it has to isotonic with blood. The concentration of glucose solution in gL-1 is __________.
(Molar mass of glucose = 180 g mol-1 R = 0.083 L bar K-1 mol-1) Nearest integer)
2.0g of H2 gas is adsorbed on 2.5g platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______________mL.
(Given : R = 0.083 L bar K-1 mol-1)
2.0g of H2 gas is adsorbed on 2.5g platinum powder at 300K and 1 bar pressure. The volume of the gas adsorbed per gram of the adsorbent is ______________mL.
(Given : R = 0.083 L bar K-1 mol-1)
Volume of H2 adsorbed =
Therefore volume of gas adsorbed per gram of the adsorbent =
1.2 mL of acetic acid is dissolved in water to make 2.0L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is_____________. (Nearest Integer).
[Given Density of acetic acid is 1.02g mL-1
Molar mass of acetic acid is 60g mol-1]
1.2 mL of acetic acid is dissolved in water to make 2.0L of solution. The depression in freezing point observed for this strength of acid is 0.0198°C. The percentage of dissociation of the acid is_____________. (Nearest Integer).
[Given Density of acetic acid is 1.02g mL-1
Molar mass of acetic acid is 60g mol-1]
Mass = d × v = 1.02 × 1.2 = 1.224gm
Moles of acetic acid = 0.0204 moles in 2L
So molality = 0.0102 mol/kg
i = 1 + a for acetic acid
0.0198 = (1 + a) × 1.85 × 0.0102
α = 0.04928 = 5%