Calculate the molar mass of the unknown solid using equations 2.2 through 2.5. Show all your work.I don’t know how to find the value for Kf either.Please help (mm)x (kg solvent) = moles solute(2.4)mass solutemoles solute= molecular mass of solute(2.5)In this experiment, you will determine the molar mass of the unknown bymeasuring the freezing-point depression observed in a solution in which the solvent islauric acid.Procedure and Collection of DataYou will watch a You Tube video in which an experiment briefly described in theintroduction is carried out. The video can be found by clicking on the following link:https://www.youtube.com/watch?v=10DrYceYwzARecord the data in the table below. Note that the temperature at the end of eachexperiment is the freezing point of the substance. There is no need to consider all ofthe time and temperature data that was collected along the way.Data Table:Mass of lauric acid (g)Mass of residual lauric acid in weighboat (g)5.0624 q43.98°0.74116q0.007140.7389Mass of lauric acid in test tube (g)190 tuld Freezing point of lauric acid (°C)villsioMass of unknown (g)Mass of residual unknown in weighboat (g)Mass of unknown in test tube (g)Freezing point of mixture (°C)38.12°cData AnalysisCalculate the molar mass of the unknown solid using equations 2.2 through 2.5. Showall of your work. General Chemistry IICHM 122Laboratory 2Determination of Molar Mass by Freezing-Point DepressionPurposeThe purpose of this experiment is to determine the molar mass of an unknowncompound by measuring the freezing-point depression observed in a solution of thecompound.Introduction cThe vapor pressure of a liquid is determined by the ability of solvent molecules tomove from the liquid phase to the gaseous phase. In a solution, however, some of thespaces at the surface of the solution are occupied by the solute particles. As a result,fewer solvent particles are in positions from which they can move into another phase,and therefore the solution's vapor pressure is lower than that of the pure solvent. Thismeans that the solution will have to be heated to a higher temperature before its vaporpressure reaches that of the atmosphere, therefore resulting in an elevated boilingpoint. Similarly, since the solution has a lower vapor pressure, the solution will have tobe cooled to a lower temperature for the solution to freeze.The extent of freezing-point depression is proportional to the amount of solutepresent in a given mass of solvent. The concentration unit employed is molality,symbolized by a lowercase m. Molality is defined as the number of moles of solute perkilogram of solvent.moles solute(2.1)m =kg solventThe difference between the freezing point of the solution and that of the pureWe cansolvent is called the freezing point depression and is given the symbol Atr.express the relationship between the difference in freezing point and the molality of thesolution in Equation 2.2:At, = k, x m(2.2)where kf is the freezing-point depression constant in units of °C-kg/mol.Equation 2.3 shows that by measuring the freezing-point depression of asolution, you can determine the solution's molality. Knowing the experimentallydetermined molality and knowing how the solution was prepared (i.e., what mass ofsolute was combined with what mass of solvent), you can obtain the molar mass ofsolute according to the following equations.At(2.3)mexpilk,

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Formulas used - 
1)  m = moles solutekg solvent
2) ∆tf = kf × m
3) mexp = ∆tfkf
4) mexp × kg solvent…