Mole Fraction Of C3H5OH3 In A Solution Of 36 G Of Water ... - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsMole fraction of ${C_3}{H_5}{(OH)_3}$ in a solution of 36 g of water and 46 g glycerine is:A. 0.46B. 0.36C. 0.20D. 0.40AnswerVerified564.3k+ viewsHint: Mole fraction of a compound is determined by first calculating the moles of the compound present in the mixture. The moles of the compound can be calculated if the mass of the compound is given by dividing it with the molecular mass of the compound.Complete step by step answer:Given,The mass of water is 36 g.Mass of glycerine is 46 g${C_3}{H_5}{(OH)_3}$ is the molecular formula of glycerine.The formula for calculating the moles is shown below.$n = \dfrac{m}{M}......(i)$Wheren is the number of moles of the compound.m is the mass of the compound.M is the molar mass of the compound.The molecular weight of water is 18.To calculate the moles of water, substitute the values of mass and molecular weight in equation (i).$n = \dfrac{{36g}}{{18g/mol}}$$\Rightarrow n = 2mol$The molecular weight of glycerine is 92 g/mol.To calculate the moles of glycerine substitute the values in the above equation.$n = \dfrac{{46g}}{{92g/mol}}$$\Rightarrow n = 0.5mol$Mole fraction of a compound is determined by dividing the moles of a compound with the total moles of the compound present in the mixture.The formula for calculating the mole fraction of a compound is shown below.${X_A} = \dfrac{{{m_A}}}{{{m_A} + {m_B}}}......(ii)$Where,${X_A}$ is the mole fraction of the compound.${m_A}$ is moles of A.${m_B}$ is moles of B.To calculate the mole fraction of glycerine, substitute the values in equation (ii).\[{X_G} = \dfrac{{0.5}}{{0.5 + 2}}\]$\Rightarrow {X_G} = 0.2$Thus, the mole fraction of ${C_3}{H_5}{(OH)_3}$ in a solution of 36 g of water and 46 g glycerine is 0.2.Therefore, the correct option is CNote:${C_3}{H_5}{(OH)_3}$ and ${C_3}{H_8}{O_3}$both are the molecular formula of glycerine. For calculating the mole fraction of water, moles of water is divided by the total moles present in the mixture.Recently Updated PagesWhy are manures considered better than fertilizers class 11 biology CBSEFind the coordinates of the midpoint of the line segment class 11 maths CBSEDistinguish between static friction limiting friction class 11 physics CBSEThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEThe first National Commission on Labour NCL submitted class 11 social science CBSENumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSEWhy are manures considered better than fertilizers class 11 biology CBSEFind the coordinates of the midpoint of the line segment class 11 maths CBSEDistinguish between static friction limiting friction class 11 physics CBSEThe Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSEThe first National Commission on Labour NCL submitted class 11 social science CBSENumber of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

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