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Sign InMole 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. 0.40

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• Class 11

• Chemistry

• Mole fraction of C3H5OH3 in a ...

AnswerStudy MaterialsNCERT Solutions For Class 11Important Questions for Class 11Revision Notes for Class 11NCERT BooksMaths Formula for Class 11Sample PapersClass 11 MathsClass 11 PhysicsClass 11 ChemistryClass 11 Business StudiesClass 11 EconomicsClass 11 AccountancySyllabusMaths SyllabusPhysics SyllabusChemistry SyllabusBiology SyllabusEnglish SyllabusHindi SyllabusAccountancy SyllabusBusiness Studies SyllabusEconomics SyllabusPolitical Science SyllabusPhysical Education SyllabusTextbook SolutionsClass 11 NCERT Exemplar SolutionsClass 11 HC Verma SolutionsClass 11 RD Sharma SolutionsClass 11 RS Aggarwal SolutionsClass 11 DK Goel SolutionsClass 11 TS Grewal SolutionsClass 11 Sandeep Garg SolutionsAnswerVerified454.8k+ 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 Pages10 Examples of Evaporation in Daily Life with Explanations10 Examples of Diffusion in Everyday Life1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSEWhat happens when dilute hydrochloric acid is added class 10 chemistry JEE_MainWhat is the meaning of celestial class 10 social science CBSEWhat causes groundwater depletion How can it be re class 10 chemistry CBSE10 Examples of Evaporation in Daily Life with Explanations10 Examples of Diffusion in Everyday Life1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSEWhat happens when dilute hydrochloric acid is added class 10 chemistry JEE_MainWhat is the meaning of celestial class 10 social science CBSEWhat causes groundwater depletion How can it be re class 10 chemistry CBSE

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