NEET Test Series from KOTA - 10 Papers In MS WORD
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CHEMISTRY(KCET)
285351
Which of the following colligative properties can provide molar mass of proteins, polymers and colloids with greater precision?
1 Depression in freezing point
2 Osmotic pressure
3 Relative lowering of vapour pressure
4 Elevation in boiling point
Explanation:
(b)
Karnataka CET 2022
CHEMISTRY(KCET)
285352
Solubility of gas in a liquid increases with
1 increase of\(P\) and decrease of \(T\)
2 decrease of\(P\) and decrease of \(T\)
3 increase of\(P\) and increase of \(T\)
4 decrease of\(P\) and increase of \(T\).
Explanation:
(a) According to Henry's law,
\(S_g=K_H P_g \quad S_g \infty P_g\)
\(S_g=\) solubility of gas
\(P_g=\) partial pressure of gas
On increasing temperature, solubility of gases decreases, as on increasing temperature, kinetic energy of gas increases.
Karnataka CET 2022
CHEMISTRY(KCET)
285353
The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is\(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is
1 \(2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
2 \(10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
3 \(0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
4 \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Explanation:
(d)\(\Delta T_b=i K_b m\)
For glucose, \(i=1\)
\(m=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
Moles of glucose \(=\frac{1.8 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}=0.01 \mathrm{~mol}\)
Mass of solvent \(=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
\(\mathrm{m}=\frac{0.01 \mathrm{~mol}}{0.1 \mathrm{~kg}}=0.1 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\(\Delta T_b=0.1=1 \times 0.1 \times K_b \Rightarrow K_b=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Karnataka CET 2022
CHEMISTRY(KCET)
285354
If 3 g of glucose (molar mass\(=180 \mathrm{~g}\) ) is dissolved in 60 g of water at \(15^{\circ} \mathrm{C}\), the osmotic pressure of the solution will be
285351
Which of the following colligative properties can provide molar mass of proteins, polymers and colloids with greater precision?
1 Depression in freezing point
2 Osmotic pressure
3 Relative lowering of vapour pressure
4 Elevation in boiling point
Explanation:
(b)
Karnataka CET 2022
CHEMISTRY(KCET)
285352
Solubility of gas in a liquid increases with
1 increase of\(P\) and decrease of \(T\)
2 decrease of\(P\) and decrease of \(T\)
3 increase of\(P\) and increase of \(T\)
4 decrease of\(P\) and increase of \(T\).
Explanation:
(a) According to Henry's law,
\(S_g=K_H P_g \quad S_g \infty P_g\)
\(S_g=\) solubility of gas
\(P_g=\) partial pressure of gas
On increasing temperature, solubility of gases decreases, as on increasing temperature, kinetic energy of gas increases.
Karnataka CET 2022
CHEMISTRY(KCET)
285353
The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is\(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is
1 \(2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
2 \(10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
3 \(0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
4 \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Explanation:
(d)\(\Delta T_b=i K_b m\)
For glucose, \(i=1\)
\(m=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
Moles of glucose \(=\frac{1.8 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}=0.01 \mathrm{~mol}\)
Mass of solvent \(=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
\(\mathrm{m}=\frac{0.01 \mathrm{~mol}}{0.1 \mathrm{~kg}}=0.1 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\(\Delta T_b=0.1=1 \times 0.1 \times K_b \Rightarrow K_b=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Karnataka CET 2022
CHEMISTRY(KCET)
285354
If 3 g of glucose (molar mass\(=180 \mathrm{~g}\) ) is dissolved in 60 g of water at \(15^{\circ} \mathrm{C}\), the osmotic pressure of the solution will be
285351
Which of the following colligative properties can provide molar mass of proteins, polymers and colloids with greater precision?
1 Depression in freezing point
2 Osmotic pressure
3 Relative lowering of vapour pressure
4 Elevation in boiling point
Explanation:
(b)
Karnataka CET 2022
CHEMISTRY(KCET)
285352
Solubility of gas in a liquid increases with
1 increase of\(P\) and decrease of \(T\)
2 decrease of\(P\) and decrease of \(T\)
3 increase of\(P\) and increase of \(T\)
4 decrease of\(P\) and increase of \(T\).
Explanation:
(a) According to Henry's law,
\(S_g=K_H P_g \quad S_g \infty P_g\)
\(S_g=\) solubility of gas
\(P_g=\) partial pressure of gas
On increasing temperature, solubility of gases decreases, as on increasing temperature, kinetic energy of gas increases.
Karnataka CET 2022
CHEMISTRY(KCET)
285353
The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is\(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is
1 \(2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
2 \(10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
3 \(0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
4 \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Explanation:
(d)\(\Delta T_b=i K_b m\)
For glucose, \(i=1\)
\(m=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
Moles of glucose \(=\frac{1.8 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}=0.01 \mathrm{~mol}\)
Mass of solvent \(=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
\(\mathrm{m}=\frac{0.01 \mathrm{~mol}}{0.1 \mathrm{~kg}}=0.1 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\(\Delta T_b=0.1=1 \times 0.1 \times K_b \Rightarrow K_b=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Karnataka CET 2022
CHEMISTRY(KCET)
285354
If 3 g of glucose (molar mass\(=180 \mathrm{~g}\) ) is dissolved in 60 g of water at \(15^{\circ} \mathrm{C}\), the osmotic pressure of the solution will be
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHEMISTRY(KCET)
285351
Which of the following colligative properties can provide molar mass of proteins, polymers and colloids with greater precision?
1 Depression in freezing point
2 Osmotic pressure
3 Relative lowering of vapour pressure
4 Elevation in boiling point
Explanation:
(b)
Karnataka CET 2022
CHEMISTRY(KCET)
285352
Solubility of gas in a liquid increases with
1 increase of\(P\) and decrease of \(T\)
2 decrease of\(P\) and decrease of \(T\)
3 increase of\(P\) and increase of \(T\)
4 decrease of\(P\) and increase of \(T\).
Explanation:
(a) According to Henry's law,
\(S_g=K_H P_g \quad S_g \infty P_g\)
\(S_g=\) solubility of gas
\(P_g=\) partial pressure of gas
On increasing temperature, solubility of gases decreases, as on increasing temperature, kinetic energy of gas increases.
Karnataka CET 2022
CHEMISTRY(KCET)
285353
The rise in boiling point of a solution containing 1.8 g of glucose in 100 g of solvent is\(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is
1 \(2 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
2 \(10 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
3 \(0.1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
4 \(1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Explanation:
(d)\(\Delta T_b=i K_b m\)
For glucose, \(i=1\)
\(m=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
Moles of glucose \(=\frac{1.8 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}=0.01 \mathrm{~mol}\)
Mass of solvent \(=100 \mathrm{~g}=0.1 \mathrm{~kg}\)
\(\mathrm{m}=\frac{0.01 \mathrm{~mol}}{0.1 \mathrm{~kg}}=0.1 \mathrm{~mol} \mathrm{~kg}^{-1}\)
\(\Delta T_b=0.1=1 \times 0.1 \times K_b \Rightarrow K_b=1 \mathrm{~K} \mathrm{~kg} / \mathrm{mol}\)
Karnataka CET 2022
CHEMISTRY(KCET)
285354
If 3 g of glucose (molar mass\(=180 \mathrm{~g}\) ) is dissolved in 60 g of water at \(15^{\circ} \mathrm{C}\), the osmotic pressure of the solution will be