What mass (kg) of the solute glycerol should be added to 3.61 L of the solvent H

Question

1. What mass (kg) of the solute glycerol should be added to 3.61 L of the solvent H2O to produce a solution that contains16.3 parts per hundred by mass glycerol?
2. What volume (mL) of the solute C6H6 should be added to 7.41 L of the solvent C4H8O to make a solution that contains 1.89 pph by mass C6H 6?
   
3. Calculate the mass (kg) of the solute (CH3)2CO and the mass (kg) of the solvent C2H5OH that should be combined to prepare 3950 g of a solution that is 1.86 parts per hundred by mass (CH 3)2CO.
4. What mass (kg) of the solvent (CH3)2CO should be added to 1190 g of the solute acetonitrile to produce a solution that is16.4% by mass acetonitrile?
5. A 4.11 kg sample of a solution of the solute methylene chloride in the solvent C2H5OH that has a concentration of28.6% by mass methylene chloride is available. Calculate theamount (kg) of CH2Cl2 that is present in the sample.
6. What mass (kg) of the solvent HCCl3 should be added to 1.91 L of the solute C6H6 to generate a solution that is 18.2 pph by mass C6H6?

Explanation / Answer

mass C4H8O = 0.8892 Kg/L x 7.41 L=6.588 Kg
1.89 = moles benzene / 6.588 Kg
moles benzene =3.48
mass benzene = 3.48 x 78.11 =272.3 g = 0.272 Kg
volume = 0.272/ 0.8892 = 0.199 L

1.86 = 1.83 moles (CH3)2CO / 1 Kg of C2H5OH

massC2H5OH = 1000 g

mass (CH3)2CO = 1.86 mol x 58.05 = 106.2 g
mass solution = 1106 g
106.2 : 1106 = x : 3950
x = mass (CH3)2CO in 3950 g of this solution =379.2 g = 0.379 Kg
V = 0.379 Kg / 789.8 Kg/m^3 = 0.0004798 m^3 => 479cm^3 => 479 mL

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