The around passes of water is 0 degree C and its K_r = 1.86 degree C/m. If we we

Question

The around passes of water is 0 degree C and its K_r = 1.86 degree C/m. If we weigh 9 grams glucose (MB - 150) and dissolve at in 50g water, the freezing point of this solution will be -1.36 degree C -1.56 degree C 0 degree C none of these freezing point depression constants of three solvents are as follows: acetic acid K_f=3.9 degree C/m: ethanol: K_f = 1.99 degree C/m: water: K_f= 1.86 degree C/m: If 3 grams of NaCI are dissolved in 30 g of each solvent, which solution has the largest depression in freezing point? Acetic acid Ethanol Water They are the same A sample of water and a sample of ethanol are both boiling in open containers at standard pressure. They are liquids have the same vapor pressure and the same temperature the same vapor pressure but different temperatures different vapor pressures but the same temperatures different vapor pressure and different temperatures. 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 affected the value of the calorimeter constant? Overstated understated no change does not apply A student is performing an experiment in which she is supposed to place 6 g of KCl into 20 mL of water, then heat the sample to find the temperature the sample dissolves. Which of the following errors will cause her temperature measurement to be lower than the actual value? The test tube is not dry before she starts the measures our more than 20 mL of water. The measure out less than 6 g of KCl. All of the above. Which of the following is a statement of Hess's Law? If a reaction is carried out in a series of steps, delta H for the reaction will equal the sum of the enthalpy changes for the individual steps. If a reaction is carries out in a series of steps, the delta H for the reaction will equal the product of the enthalpy changes for the individual steps. The delta H of a reaction depends on the physical states of the reactants and products. A pressure of 1.0 atm is the same as a pressure of __ of mm Hg. 193 760.0 33.0 101 29.92

Explanation / Answer

30. Given normal freezing point of water=0oC

Kf for water=1.86 oC/m

Mass of glucose=9 g

Molar mass of glucose=180 g/mol

Mass of solvent (water)=50 g

Molality of solution=number of moles of glucose/mass of water (kg)

=(mass of glucose/Molar mass of glucose)/Mass of water (kg)

=(9 g/180 g)/50/1000 g/kg

=1 m

Glucose is a non-electrolytic solute

Depression in freezing point=Kfm

=(1.86 oC/m)(1 m)

=1.86 oC

So the freezing point of the solution=0o C-1.86 oC=-1.86 oC i.e. option (a)

31. Depression is freezing point=iKfm

i=van't hoff factor (depeds upon solute)

m=molality of solution (depends upon mass of solute and mass of solvent)

Kf=Freezing point depression constant (depends upon solvent)

For a given mass of the same solute and for a given mass of different solvents both i and m will be same in all cases. So depression in freezing point will depend upon the solvent i.e. on the Kf value. Since Acetic acid has the largest value for Kf, so it will have the largest depression in freezing point. Som option (a) is the answer.

32. At boiling point, vapour pressure of a liquid becomes equal to atmospheric pressure. Since both the samples of water and ethanol are boiling so that means they have the same vapour pressure equal to the atmospheric pressure. But since water and ethanol have different boiling points, so the two liquids will be at different temperatures while boiling. So option (b) is correct, that is they have same vapour pressures but different temperatures.

33. In calculation of calorimeter constant, when we transfer hot water into cold water we take the initial and final temperatures for both hot and cold water and we calculate the heat gained by cold water and heat lost by cold water

Heat gained by cold water=Rise in temperaturexSpecific heat of waterxMass of cold water

Heat lost by hot water=Decrease in temperaturexSpecific heat of waterxMass of hot water

Then we subtract the energy gained by the cold water from the energy lost by the hot water. This gives the energy gained by the calorimeter. We then divide this value by the Rise in temperature of the cold water. Thus if we do not pour all the weighted hot water into the cold water, the value of the energy gained by the calorimeter will come out to be more than it actually is (because the weight of hot water actually consumed is less than the weight of the hot water measured) thus the calorimeter constant will also come out to be more than it actually is that is it is overstated. So option (a) is correct.

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