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2 years ago

A solution that contains 55.0 g of ascorbic acid (vitamin C) in 250.0 g of water freezes at –2.34°C.

Chemistry

1 answer:

Ksju [112]2 years ago

3 0

Answer:

174.8 g/m is the molar mass of the solute

Explanation:

We must apply colligative property of freezing point depression.

ΔT = Kf . m . i

ΔT = T° freezing pure solvent - T° freezing solution (0° - (-2.34°C) = 2.34°C

Kf = Fussion constant for water, 1.86 °C/m

As ascorbic acid is an organic compound, we assume that is non electrolytic, so i = 1

2.34°C = 1.86°C/m . m

2.34°C / 1.86 m/°C = 1.26 m

This value means the moles of vitamin C, in 1000 g of solvent

We weighed the solute in 250 g of solvent, so let's calculate the moles of vitamin C.

1000 g ___ 1.26 moles

In 250 g ___ (250 . 1.26)/1000 = 0.314 moles

This are the moles of 55 g of ascorbic acid, so the molar mass, will be:

grams / mol ⇒ 55 g/0.314 m = 174.8 g/m

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If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the

Kobotan [32]

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

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1 year ago

How many atoms are in a molecule of RSq?

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2 atoms in a molecule of RSq, TSq contains 3 different types of atoms (Sq, R, and T.)

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Calculate the percent composition for each of the elements in Na3PO4. A 5.00 gram sample of an oxide of lead PbxOy contains 4.33

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Answer:

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(ii) 2C₂H₆ + 7O₂ ---> 4CO₂ + 6H₂O

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4. 48 g of MG will react with 2 moles of Cl₂

5. 0.288 g of SO2 will be produced from the combustion of 0.331 g P₄S₃ in excess O₂

6. 12.8 g of nitric oxide can be produced from the reaction of 8.00 g NH₃ with 17.0 g O₂

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Explanation:

The full explanation is found in the attachments below

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2 years ago

Which statement below correctly identifies the number of subatomic particles inside and outside the nucleus of an argon atom?

Juliette [100K]

Answer:

Explanation:

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2 years ago

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

ratelena [41]

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial: 0.500 0.500 0.005 0.005

At eqllm: 0.500-x 0.500-x 0.005+x 0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

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2 years ago