A solution is prepared by mixing 1 00 p ethanol (C_2H_5OH) with 100.0 f water to

Question

A solution is prepared by mixing 1 00 p ethanol (C_2H_5OH) with 100.0 f water to five a final volume of 101 mL Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution A certain soft dank is bottled so that a bottle at 25 degree C contains CO_2 gas at a pressure of 5 0 arm over the liquid Assuming that the partial pressure of CO_2 in the atmosphere is 4 0 times 10^-4 atm. calculate die equilibrium concentrations of CO_2 in the soda both before and after the bottle is opened The Henry's law constant for CO; m aqueous solution is 3 1 times 10^-2 mol. L middot atm at 25 degree C A solution is prepared by mixing 5 81 g acetone (C_3H_6o. molar mass 58 1 g/mol.) and 11.9 f chloroform (HCCl_3, molar mass = 119.4 0m1). At 35 degree C, this solution has a total vapor pressure of 260 ton Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35 degree C are 345 and 293 torr, respectively.

Explanation / Answer

25. Given mass of C2H5OH=1g , mass of H2O=100g , Volume of solution = 101 mL

To find : molarity, mass percent, mole fraction and molality of ethanol

Molarity(M) is the number of gram moles of solute dissolved per litre of solution

Molarity = gram moles of solute / volume of solution in litre

        M=(mass of solute in grams/molar mass)/(volume of solution in mL/1000)

molar mass of C2H5OH=2*12+5*1+1*16+1*1 = 24+5+16+1=46 g/mol

Atomic mass of C=12,H=1,O=16

        M=(1/46)/(100/1000) = 1 * 1000/46* 100 =10/46 = 0.217 molL-1(or 0.217M)

26. Given is p=5 atm, pCO2=4*10-4atm, Henry's constant kH=3.1*10-2mol/L

To find: concentration of CO2 before & after

According to Henry's law mass of gas dissolved in a given volume of liquid depends on the pressure applied at constatnt temperature.

so m =kH * p

that is solubility = kH* p

solubility of gas before the bottle is opened = kH * p=(3.1* 10-2) * 5 = 0.16 M

Solubility of gas after the bottle is opened = kH * p=(3.1 * 10-2) * (4 * 10-4)= 1.2 * 10-5 M

27.Given is mass of acetone C3H6O=5.81g , Molar mass of C3H6O=58.1g/mol

                 mass of chloroform CHCl3=11.9 g,Molar mass of CHCl3=119.4g/mol

Total vapour pressure P=260 torr , vapour pressure of C3H6O pC3H6O=345 torr, vapour pressure of CHCl3 pCHCl3= 293torr

no of moles of acetonel nC3H6O= mass/molar mass =5.81/58.1=0.1 mol

no of moles of chloroform nCHCl3=mass/molar mass=11.9/119.4=0.1 mol

So nC3H6O=nCHCl3=0.1 mol

mole fraction xC3H6O=nC3H6O/(nC3H6O+nCHCl3) =0.1/(0.1+0.1)=0.500 mol

mole fraction xCHCl3=nCHCl3/(nCHCl3+nC3H6O)=0.1/(0.1+0.1)=0.500 mol

According to Raoult's Law pA=poA * xA

For ideal solution Ptotal=pAxA+pBxB =pC3H6OxC3H6O+ pCHCl3xCHCl3

                                  =345*0.5 + 293* 0.5 =319 torr

Given total vapour pressure is 260 torr and calculated vapour pressure is 319 torr.Hence this is not an ideal solution

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