Ask question

Ask question

All categories

2 years ago

12

In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc

entration of this solution?

1 answer:

Hatshy [7]2 years ago

3 0

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = $\frac{0.325}{1.86}$

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

You might be interested in

12. The vapor pressure of water at 90°C is 0.692 atm. What is the vapor pressure (in atm) of a solution made by dissolving 3.68

luda_lava [24]

Answer : The vapor pressure (in atm) of a solution is, 0.679 atm

Explanation : Given,

Mass of $H_2O$ = 1.00 kg = 1000 g

Moles of $CsF$ = 3.68 mole

Molar mass of $H_2O$ = 18 g/mole

Vapor pressure of water = 0.692 atm

First we have to calculate the moles of $H_2O$ .

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{1000g}{18g/mole}=55.55mole$

Now we have to calculate the mole fraction of $H_2O$

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CsF}=\frac{55.55}{55.55+3.68}=0.938$

Now we have to partial pressure of solution.

According to the Raoult's law,

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

where,

$P_{Solution}$ = vapor pressure of solution

$P^o_{H_2O}$ = vapor pressure of water = 0.692 atm

$X_{H_2O}$ = mole fraction of water = 0.938

$P_{Solution}=X_{H_2O}\times P^o_{H_2O}$

$P_{Solution}=0.938\times 0.692atm$

$P_{Solution}=0.649atm$

Therefore, the vapor pressure (in atm) of a solution is, 0.679 atm

5 0

1 year ago

Caffeine, a molecule found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 10.0 g of caffeine pr

denis-greek [22]

Answer:

194 g/mol.

Explanation:

Hello,

In this case, one first must compute the mass of each element as shown below:

$C=18.13gCO_2*\frac{12gC}{44gCO_2} =4.945gC\\H=4.639gH_2O*\frac{2.016gH}{18.0152gH_2O}=0.519gH\\N=2.885gN_2\\O=10.0g-4.945g-0.519g-2.885g=1.651gO$

Next, the corresponding moles:

$C=4.945gC*\frac{1molC}{12gC}=0.412mol\\H=0.519gH*\frac{1molH}{1gH}=0.519mol\\N=2.885gN*\frac{1molN}{14gN}=0.206molN\\O=1.648gO*\frac{1molO}{16gO} =0.103molO$

Then, each element's subscripts is found to be:

$C=\frac{0.412}{0.103}=4\\H=\frac{0.519}{0.103}=5\\N=\frac{0.206}{0.103} =2\\O=\frac{0.103}{0.103}=1$

Therefore, the empirical formula is:

$C_4H_5N_2O$

Nonetheless, it has a molar mass of 97bg/mol, thereby, by multiplying such formula by 2 one gets:

$C_8H_10N_4O_2$

Which has a molar mass of 194 g/mol being correctly contained in the given interval.

Best regards.

8 0

2 years ago

The equilibrium reaction below has the Kc = 4.00 at 25°C. If the temperature of the system at equilibrium is increased to 100°C,

professor190 [17]

Answer:

Following are the answer to this question:

Explanation:

In the given question information is missing, that is equation which can be defined as follows:

$CH_4(g)+H_2O(g)\rightarrow CO(g)+3H_2(g) \ \ \bigtraingleup H^0=+206.2KJ$

Growing temperatures may change its connection to just the way which consumes thermal energy in accordance with Le chatelier concepts Potential connection is endothermic. Answer: shifts to the right

Kc are described as a related to the concentration by the intensity of both the reaction for each phrase which reaches a power equal towards its stoichiometric equation coefficient Kc = \frac{product}{reactant} It increases [product] but reduces [reactant] Therefore, Kc increases

6 0

2 years ago

Select all that reasons that the reaction contents does not conduct at this low point.

Alexxx [7]

It's all three of the answers

5 0

1 year ago

What is the concentration of Iodine I2 molecules in a solution containing 2.54 g of iodine 250 cm3 of solution? A 0.01mol/dm3 B

gulaghasi [49]

Answer:

C. 0.04 moles per cubic decimeter.

Explanation:

The molar mass of the Iodine is 253.809 grams per mole and a cubic decimeter equals 1000 cubic centimeters. The concentration of Iodine ( $c$ ), measured in moles per cubic decimeter, can be determined by the following formula:

$c = \frac{m}{M\cdot V}$ (1)

Where:

$m$ - Mass of iodine, measured in grams.

$M$ - Molar mass of iodine, measured in grams per mol.

$V$ - Volume of solution, measured in cubic decimeters.

If we know that $m = 2.54\,g$ , $M = 253.809\,\frac{g}{mol}$ and $V = 0.25\,dm^{3}$ , then the concentration of iodine in a solution is:

$c = \frac{2.54\,g}{\left(253.809\,\frac{g}{mol} \right)\cdot (0.25\,dm^{3})}$

$c = 0.04\,\frac{mol}{dm^{3}}$

Hence, the correct answer is C.

3 0

2 years ago