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This is the second of our many quizzes on Colligative properties. Click on this to access more such competitive quizzes on Physical Chemistry.

Colligative properties Quiz 2

chemistry

  In the laboratory equal volume of 0.30 M  $BaC{{l}_{2}}$ and 0.30 M $N{{a}_{2}}S{{O}_{4}}$ are mixed together. The concentration of chloride ion in the mixture will be:

 To solve this question we have to find the moles of chloride first and then the concentration.

  In the question equal volume of $BaC{{l}_{2}}$ and $N{{a}_{2}}S{{O}_{4}} \\$ is given therefore we will take volume as 1L. Now, we have to find the concentration of chloride ion in the mixture. 1 L of 0.30 M $BaC{{l}_{2}}$ contains 2 chlorides ion. So, moles of chloride ion = $\text{molarity} \times \text{volume}$. Therefore, moles = 0.60. So, the molarity or concentration of chloride ion = $\dfrac{\text{moles} }{\text{volume}}\\ \therefore Molarity=\dfrac{0.60}{1}=0.60$ 

 Among the following, choose the compounds that shows a non-ideal solution having positive deviation from Raoult's Law:

 The solution with high vapor pressure shows positive deviation from Raoult's law and the intermolecular attraction between the compounds will be reduced.

  Ethanol and cyclohexane show positive deviation from the Raoult's law because in ethanol the molecules are held together by hydrogen bonding and when we add cyclohexane to ethanol the molecules of the cyclohexane tries to occupy the space between the ethanol molecule due to which some hydrogen bonds are break. Therefore the intermolecular attraction between ethanol and cyclohexane is also reduced and the solution shows higher vapor pressure.

 Freezing point of camphor is $179{}^\circ C$. After the addition of $1.3$ g of non-volatile solute in $50$ g of camphor the freezing point lowers to $130{}^\circ C$. Find the molecular weight of the non-volatile solute added to the camphor: $\left( {{K}_{f}}(camphor)=39.70KKgmo{{l}^{-1}} \right)$

 To solve this question you have to use the formula for depression in freezing point. Change the units of temperature.

 From your chemistry lesson you have learnt about the depression in freezing point which is given as $\Delta {{T}_{f}}=\dfrac{{{K}_{f}}\times {{w}_{2}}\times 1000}{{{M}_{2}}\times {{w}_{1}}}$ Now, $\Delta {{T}_{f}}=403-452=-49K.$ So, the expression to find the molecular weight of the solute will be: ${{M}_{2}}=\dfrac{{{K}_{f}}\times {{w}_{1}}\times 1000}{\Delta {{T}_{f}}\times {{w}_{1}}}.$ After substituting all the value in the above expression we will get, ${{M}_{2}}=\dfrac{39.70\times 1.3\times 1000}{-49\times 50}$ = -21.0g/mol. Therefore the molecular weight of the solute will be -21.0 g/mol.

 Among the following choose the best solvent which is used for the determination of solute by depression in freezing point.

 The molal depression constant or the cryoscopic constant will be high for those solvent which is used for the determination of solute by depression in freezing point.

 Among the following solvent given in the option camphor is the best solvent which is used for the determination of solute by depression in freezing point because the molal depression constant $({{K}_{f}})$ of camphor is very high.

  In three different containers three types of macromolecules such as polymers, colloids and protein are kept. Choose the correct property which is used for the determination of molar mass of these three macromolecules.

 The property which is used for the determination of molar mass of these three macromolecules is a colligative property and also there is no temperature change involved in this property.

Osmotic pressure is the best method which  is used for the determination of molar mass such as polymers, colloids and protein macromolecules. Because as we know that osmotic pressure involves no temperature change and it is a colligative property  therefore there is no chance of change occurring in nature these macromolecules.