Question

1a) In lab, a student mixes 30.8001 mL of benzene in 177mL of hexane, what is the MOLALITY of this solution?

benzene, density= 0.877 g/mL

hexane, density = 0.660 g/mL

1b) A student in during lab wishes to calculate the MOLARITY, MOLALITY, and the MOLE FRACTION of NH3 in a 15.201 mass % aqueous solution (d= 0.9651 g/mL)

1c) When creating sherbert, all of the ingredients are kept below 0.0*C in an ice-salt bath. You can assume that NaCl dissolves comoletely and forms an ideal solution. What MASS is required to bring the melting point of 7.50 kg of ice to -5.0*C ? (Kf H2O =1.86*C/m)

Please thoroughly explain all of your answers. I attenpted to do all of these and had the wrong answer. I want to understand. Thank you!

Answer 1

1a) In lab, a student mixes 30.8001 mL of benzene in 177mL of hexane, what is the MOLALITY of this solution?

benzene, density= 0.877 g/mL   ==> mass of benzene = 30.8001 mL*0.877 g/mL = 27.01 g

moles of Benzene = 27.01 g / 78 g/mol = 0.3463 moles

(molar mass of Benzene = 78 g/mol)

hexane, density = 0.660 g/mL    ==> mass of Hexane = 177mL* 0.660 g/mL     = 116.82 g

MOLALITY of this solution , m = moles of Solute / mass of solvent ( in Kg)

m = 0.3463 mole / 0.11682 Kg = 2.9644 m

1b) A student in during lab wishes to calculate the MOLARITY, MOLALITY, and the MOLE FRACTION of NH3 in a 15.201 mass % aqueous solution (d= 0.9651 g/mL)

NH3 in a 15.201 mass % aqueous solution

Density of solution : 0.9651 g/mL

thus mass of solution per L : 965.1 g/L

mass of NH3 in aqueous solution = 146.7 g /L

molarity of NH3 solution: 146.7 g/L / 17 g/mol = 8.63 mol/L or 8.63 M

mass of solvent : 965.1 g - 146.7 g = 818.4 g or 0.8184 Kg

thus, molality : 8.63 mol/ 0.8184 Kg = 10.545 m

MOLE FRACTION

: moles of water : 818.4 g / 18 g/mol = 45.47 moles

mole fraction , x = 8.63 mol / ( 8.63 + 45.47) mol = 0.16

1c) When creating sherbert, all of the ingredients are kept below 0.0*C in an ice-salt bath. You can assume that NaCl dissolves completely and forms an ideal solution. What MASS is required to bring the melting point of 7.50 kg of ice to -5.0*C ? (Kf H2O =1.86*C/m)

we have , depression in freezing point :

Tf = i*Kf * m

for NaCl , i = 2

5.0 C = 2*1.86 C/m * m

molality = 1.344 m

molality = moles of solute / solvent (in Kg )

moles of solute / 7.50 Kg = 1.344 m

moles of solute = 10.08 moles

mass of NaCl required : 10.08 moles* 58.5 g/mol = 589.6 g