All categories

2 years ago

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.

Chemistry

1 answer:

Rainbow [258]2 years ago

4 0

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

You might be interested in

What would happen if carbon -14 gained 1 proton and lost 1 neutron?

sleet_krkn [62]

D. To answer this question refer to the periodic table and think logically about it. Adding one proton increases the atomic number so you go along the row to Nitrogen. If you lose one neutron then the atomic mass decreases by 1 so 14-1 is 13.

4 0

3 years ago

How many water molecules self ionize in one liter of water

Drupady [299]

Answer:

yes

Explanation:

because I think it is correct fo u

5 0

3 years ago

What is the total number of atoms in the molecule

serg [7]

Answer:

Sulfuric acid contains 2 hydrogen atoms, 1 sulfur atom, and 4 oxygen atoms.

Explanation:

8 0

2 years ago

An analytical chemist weighs out 0.188 g of an unknown triprotic acid into a 250 mL volumetric flask and dilutes to the mark wit

Anon25 [30]

<u>Answer:</u> The molar mass of unknown triprotic acid is 97.66 g/mol

<u>Explanation:</u>

To calculate the molarity of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of triprotic acid

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=3\\M_1=?M\\V_1=250mL\\n_2=1\\M_2=0.0600M\\V_2=95.9mL$

Putting values in above equation, we get:

$3\times M_1\times 250=1\times 0.0600\times 95.9\\\\M_1=0.0077M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = 0.0077 M

Given mass of triprotic acid = 0.188 g

Volume of solution = 250 mL

Putting values in above equation, we get:

$0.0077M=\frac{0.188\times 1000}{\text{Molar mass of triprotic acid}\times 250}\\\\\text{Molar mass of triprotic acid}=97.66g/mol$

Hence, the molar mass of unknown triprotic acid is 97.66 g/mol

7 0

2 years ago

How could you verify that you produced carbon dioxide in your combustion reaction? 2. what indication did you have that nh3 was

DanielleElmas [232]

1) Carbon dioxide is a gas, so when $CO_{2}$ is evolved in the reaction, it appears as bubbles. The gas released extinguishes the fire and it can turn lime water milky.

$Ca(OH)_{2} (aq) + CO_{2}(g) ----> CaCO_{3}(s) +H_{2}O(l)$

2) When $NH_{3}$ is released in a decomposition reaction we can identify by the strong pungent smell of the gas released.

3) Saturated citric acid can cause corrosion of the metal layers present in the pipes. So, before draining out any acid it is neutralized so that the pipes and other plumbing works do not get damaged leading to leaks in the drainage system.

5 0

3 years ago