onnaitreezing point ofwater is 0 and its Kr-1.86°C/ m. If we weigh 9 grams gluco

Question

onnaitreezing point ofwater is 0 and its Kr-1.86°C/ m. If we weigh 9 grams glucose (MW 29. The -180) and dissolve it in 50 g water, the freezing point of this solution will be (c) 0 (a) +1.86 (b)-1.86 (d) none of these Molar freezing point depression constants of three solvents are as follows: acetic acid: Kr=3.9/ m ; ethanol: K.-199/ m ; water: K-186°C/m; If 3 in rensi of NaCi are dissolved in 30 g of each solvent, which solution has the largest depression freezing point? (a) Acetic acid (b) Ethanol (e) Water (d) They are the same 31. When the vapor pressure of a liquid in an open container equals the atmospheric pressure, the liquid will c. melt d. crystallize a. freeze b. boil 32. When determining the calorimeter constant for a styrofoam cup calorimeter, a student did not pour all the weighted hot water into the cool water. How will this affect the value of the calorimeter constant? a.) overstated b) understated c)no change d) does not apply Under the same conditions of temperature and pressure, a liquid differs from a gas because the 33. particles of the liquid a. take the shape of the container they occupy b. have no regular arrangement c. have stronger forces of attraction between them 34. As the temperature of a liquid increases, its vapor pressure a.) increases b.) decreases c.) remains the same A pressure of 1.0 atm is the same as a pressure of mm Hg. 35. e.33.0 a. b. 193 101 c. 760.0 d. 29.92

Explanation / Answer

29)

Lets calculate molality first

mass of solute = 9 g

we have below equation to be used:

number of mol of solute,

n = mass of solute/molar mass of solute

=(9.0 g)/(180 g/mol)

= 5*10^-2 mol

mass of solvent = 50 g

= 5*10^-2 kg [using conversion 1 Kg = 1000 g]

we have below equation to be used:

Molality,

m = number of mol / mass of solvent in Kg

=(5*10^-2 mol)/(0.05 Kg)

= 1 molal

lets now calculate deltaTf

deltaTf = Kf*m

= 1.86*1

= 1.86 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 1.86

= -1.86 oC

Answer: b

Only 1 question at a time please

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.