Question

A solution of a sugar with the chemical formula (c12h22o11) is prepared by dissolving 8.45 g in 250.0 ml of water at 25

Answer 1

A. Concentration in terms of molarity

First, calculate for the molar mass of the given solution, C12H22O11.
    m = 12(12) + 22(1) + 11(16) = 232
Calculate for the number of moles.
   
    n = 8.45 g / 232 g/mol  = 0.036422 moles

Divide the determined number of moles by volume of solution in L.
    V(solution) = (250 mL)(1 L/1000 mL) = 0.25 L

The concentration in molarity is determined by dividing the number of moles by the volume of solution in liters.
          M = 0.03622 moles / 0.25 L = 0.146 moles/L

b. Concentration in terms of molality
Calculate the mass of the solvent in kilogram given the volume and its density.
              m(solvent) = (1.21 g/cm3)(250 mL)(1 kg/1000 g) = 0.3025 kg
Divide the calculated number of moles by the mass of the solvent in kilograms.

         molality = 0.036422 moles/0.3025 kg = 0.12 molal

c. Concentration in terms of weight percent.
The total weight of the solution is equal to 302.5 g.
    
           weight percent of solute = (8.45 g / 302.5 g)(100%) = 2.79%

d. Weight percent in ppm.
           weight percent of solute x (10000 ppm/1%) 
       = (2.79%)(10000 ppm/1%) = 27933.88 ppm

Answer 2

Answer : The molarity, molality, weight percent and ppm are, 0.0987 mole/L, 0.0839 mole/Kg, 2.79% and 27933.8843 respectively.

Solution : Given,

Mass of solute $(C_{12}H_{22}O_{11})$ = 8.45 g

Molar mass of solute $(C_{12}H_{22}O_{11})$ = 342.3 g/mole

Volume of solvent (water) = volume of solution = 250 ml

Density of solution = $1.21g/cm^3=1.21g/ml$

Calculation for molarity

$Molarity=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{volume of solution in ml}}$

$Molarity=\frac{8.45g\times 1000}{342.3g/mole\times 250ml}=0.0987mole/L$

Now we have to calculate the mass of solution.

$\text{Mass of solution}=Density\times Volume=1.21g/ml\times 250ml=302.5g$

Now we have to calculate the mass of solvent.

$\text{Mass of solvent}=\text{Mass of solution}-\text{Mass of solute}=302.5-8.45=294.05g$

Calculation for molality

$Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent in g}}$

$Molality=\frac{8.45g}\times 1000}{342.3g/mole\times 249.05g}=0.0839mole/Kg$

Calculation for weight percent

$weight\%=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 100=\frac{8.45g}{302.5g}\times 100=2.79\%$

Calculation for ppm

$ppm=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times 10^6=\frac{8.45g}{302.5g}\times 10^6=27933.8843$