","comment":{"@type":"Comment","text":" This type of solutions obeys Raoult's law and the total vapor pressure of the solution is given as ${{P}_{s}}={{P}_{A}}+{{P}_{B}}$."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Non–Ideal solution","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" Elevation in boiling point","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" Non-Ideal solution showing positive deviation","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" Ideal solution","position":2,"answerExplanation":{"@type":"Comment","text":" As we know from the chemistry lessons that the solutions for the mixture of a volatile liquids which obeys the Raoult's law at all the temperature and concentrations is known as an ideal solution and the total vapor pressure of the solution is given as $${{P}_{s}}={{P}_{A}}+{{P}_{B}}.$$ Example of an ideal solution is benzene and toluene.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Colligative properties Quiz 1","text":" What amount of sodium chloride will be added to six liters of water if the freezing point of solution is $$-5.6{}^\\circ C$$ and the density of water is 1g/ml. $$\\left( {{K}_{{{f}_{{{H}_{2}}O}}}}=2.65\\dfrac{{}^\\circ C}{m} \\right)$$ ","comment":{"@type":"Comment","text":" To solve this question we have to use the formula of depression in freezing point that is given by $\\Delta {{T}_{f}}={{K}_{f}}\\times i\\times m$ where i is the van't Hoff factor for NaCl."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 79g","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 157g","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 169g","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 370g","position":0,"answerExplanation":{"@type":"Comment","text":" From the chemistry lessons we have learnt about the depression in freezing point and it is given by $$\\\\ \\Delta {{T}_{f}}={{K}_{f}}\\times i\\times m.\\\\$$ Where, $$\\Delta {{T}_{f}}= \\text{Depression of freezing point}= T_{f}^{o}-{{T}_{f}} [T_{f}^{o}= \\text{freezing point of pure solvent} (0^oC) \\\\ \\text{and} {{T}_{f}} \\text{is the freezing point of the solution)}\\\\ \\text{i = van't hoff factor (For Nacl it is 2)}\\\\ \\text{Therefore, from the question} \\Delta {{T}_{f}}= 0^oC- (-5.6{}^\\circ C) \\\\ =5.6{}^\\circ C \\\\ \\text{Now, molality} = \\dfrac{{{W}_{2}}\\times 1000}{{{M}_{1}}\\times {{W}_{1}}}\\\\ \\therefore \\Delta {{T}_{f}}={{K}_{f}}\\times i\\times \\dfrac{{{W}_{2}}\\times 1000}{{{M}_{1}}\\times {{W}_{1}}}\\\\ \\text{After putting all the values in the formula we get:}\\\\ 5.6=2.56\\times 2\\times \\dfrac{{{W}_{2}}\\times 1000}{58.5\\times 6000}=370g\\\\ \\text{(weight of water }= 6000g \\text{(d=m/v)} \\\\ \\text{and molecular weight of NaCl is } 58.5)$$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Colligative properties Quiz 1","text":" Calculate the ideal van't Hoff factor (i) value for the given compounds: $$HN{{O}_{3}},Mg{{(Cl{{O}_{3}})}_{2}},\\,NaCl,\\,AgN{{O}_{3}}$$","comment":{"@type":"Comment","text":" To calculate the van't Hoff factor of the given compounds we have to calculate the no. of particles present in it."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 3, 4, 2, 1","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 1, 2, 2, 2","position":2},{"@type":"Answer","encodingFormat":"text/html","text":" 2, 1, 1, 2","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 2, 3, 2, 2","position":0,"answerExplanation":{"@type":"Comment","text":" The total number of particles in $$HN{{O}_{3}},Mg(Cl{{O}_{3}}),NaCl,AgN{{O}_{3}}$$ will be: $$ \\\\ HN{{O}_{3}}\\to {{H}^{+}}+NO_{3}^{-}\\ \\text{(i=1+1=2)} \\\\ Mg{{(Cl{{O}_{3}})}_{2}}\\to M{{g}^{2+}}+2ClO_{3}^{-} \\text{(i=2+1=3)} \\\\ NaCl\\to N{{a}^{+}}+C{{l}^{-}} \\text{(i=1+1=2)} \\\\ AgN{{O}_{3}}\\to A{{g}^{+}}+NO_{3}^{-} \\text{(i= 1+1=2)} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Colligative properties Quiz 1","text":" Find the van't Hoff factor of the given aqueous solutions and arrange them in the order of their decreasing freezing points. (a) $$0.60\\,m\\,AgC{{l}_{2}}$$ (b) $$0.20\\,m\\,{{H}_{2}}S{{O}_{4}}$$ (c) $$0.30\\,m\\,NaCl$$ (d) $$0.40\\,m\\,MgC{{l}_{2}}$$","comment":{"@type":"Comment","text":" Larger the concentration of the compound smaller will be the freezing point. $\\Delta {{T}_{f}}={{K}_{f}}\\times i\\times m$ $(\\text{i= van't Hoff factor})$"},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" a > b > c > d","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" b > d >c >a","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" d > a > b = c","position":2}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" b = c > d > a","position":3,"answerExplanation":{"@type":"Comment","text":" Smaller the freezing point larger will be the concentration of the compound. $$ \\\\ \\Delta {{T}_{f}}={{K}_{f}}\\times i\\times m(\\text{i= van't Hoff factor})\\\\ \\text{For } 0.60\\,m\\,AgC{{l}_{2}}\\\\ AgC{{l}_{2}}\\to A{{g}^{2+}}+C{{l}^{-}}(i=2+1=3)\\\\ \\Delta {{T}_{f}}=0.60\\times 3=1.80\\\\ \\text{For} 0.20\\,m\\,{{H}_{2}}S{{O}_{4}} \\\\ {{H}_{2}}S{{O}_{4}}\\to {{H}^{+}}+SO_{4}^{2-} (i= 1+2=3) \\\\ \\Delta {{T}_{f}}=0.20\\times 3=0.60\\\\ \\text{For} 0.30\\,m\\,NaCl\\ NaCl\\to N{{a}^{+}}+C{{l}^{-}} (i= 1+1=2)\\\\ \\Delta {{T}_{f}}=0.30\\times 2=0.60\\\\ \\text{For} 0.40\\,m\\,MgC{{l}_{2}}\\\\ MgC{{l}_{2}}\\to M{{g}^{2+}}+C{{l}^{-}} (i= 2+1=3)\\\\ \\Delta {{T}_{f}}=0.40\\times 3=1.20\\\\ \\therefore 0.20m\\,{{H}_{2}}S{{O}_{4}} >0.30m\\,NaCl >\\,0.40m\\,MgC{{l}_{2}} >\\,0.60mAgC{{l}_{2}} $$","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]},{"@type":"Question","eduQuestionType":"Multiple choice","learningResourceType":"Practice problem","educationalLevel":"beginner","name":"Colligative properties Quiz 1","text":" In the question some of the solutions such as, $$NaCl,\\,BaC{{l}_{2}},\\,MgS{{O}_{4}},\\,Urea,\\,Ba{{(N{{O}_{3}})}_{2}}\\,and\\,{{H}_{2}}S{{O}_{4}}$$ are given with their respective concentration. Find the pair of solutions that are isotonic at the same temperature?","comment":{"@type":"Comment","text":" Isotonic solutions are those types of solutions which have same osmotic pressure and at same temperature if the osmotic pressure is same the concentration must also be equal."},"encodingFormat":"text/markdown","suggestedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 0.3 M urea and 0.3 M $$NaCl$$ ","position":0},{"@type":"Answer","encodingFormat":"text/html","text":" 0.1 M $$MgS{{O}_{4}}$$ and 0.1 M $$Ba{{(N{{O}_{3}})}_{2}}$$ ","position":1},{"@type":"Answer","encodingFormat":"text/html","text":" 0.4 M $$BaC{{l}_{2}}$$ and 0.3 M urea","position":3}],"acceptedAnswer":[{"@type":"Answer","encodingFormat":"text/html","text":" 0.2 M $${{H}_{2}}S{{O}_{4}}$$ and 0.2 M $$BaC{{l}_{2}}$$ ","position":2,"answerExplanation":{"@type":"Comment","text":"Osmotic pressure is given as $$\\pi =iCRT$$, where $$\\pi$$ is the osmotic pressure , i is the van't Hoff factor and C is the concentration. For 0.2 M $${{H}_{2}}S{{O}_{4}}$$ and 0.2 M $$BaC{{l}_{2}}$$, the osmotic pressure will be same because $$\\\\ {{H}_{2}}S{{O}_{4}}\\to {{H}^{+}}+SO_{4}^{2-}\\ \\text{(i= 1+2=3)}. \\\\$$ So, $$\\pi =3\\times 0.1RT=0.3RT\\\\ BaC{{l}_{2}}\\to B{{a}^{2+}}+C{{l}^{-}} (i= 2+1=3). \\\\$$ So, $$\\pi =3\\times 0.2RT=0.6RT.\\\\$$ Therefore, as the osmotic pressures are equal they will be isotonic.","encodingFormat":"text/html"},"comment":{"@type":"Comment"}}]}]}