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Zepler [3.9K]
3 years ago
6

An aqueous solution of methanol (MM = 32.04 g/mol) has a molality of 8.83 m and a density of 1.15 g/mL. What is the molarity of

methanol in the solution?
Chemistry
1 answer:
hodyreva [135]3 years ago
7 0

We have that the  the molarity of methanol in the solution is mathematically given as

Morality=7.91mol/l

<h3>Chemical Reaction</h3>

Question Parameters:

An <u>aqueous </u><em>solution </em>of methanol (MM = 32.04 g/mol)

A molality of 8.83 m and a density of 1.15 g/mL.

Generally the equation for the   is mathematically given as

Morality=\frac{moles of solute }{mass of the solvent}

where

Mass of solution=(100+282.2)

Mass of solution=1289.9

Volume=1289.9/1.15

Volume=1.115L

Therefore

Morality=\frac{moles of solute }{mass of the solvent}

Morality=8.83/1.115

Morality=7.91mol/l

For more information on Chemical Reactionvisit

brainly.com/question/11231920

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4 years ago
a compound has an empirical formula of CH2 what is the molecular formula if it's molar mass is 252.5 grams/mol
Goryan [66]

Empirical formula is the simplest ratio of components making up the compound. the molecular formula is the actual ratio of components making up the compound.

the empirical formula is CH₂. We can find the mass of CH₂ one empirical unit and have to then find the number of empirical units in the molecular formula.

Mass of one empirical unit - CH₂ - 12 g/mol x 1 + 1 g/mol x 2 = 12 = 14 g

Molar mass of the compound is - 252 .5 g/mol

number of empirical units = molar mass / mass of empirical unit

                                           = \frac{252.5 g/mol}{14 g}

                                           = 18 units

Therefore molecular formula is - 18 times the empirical formula

molecular formula  - CH₂ x 18 = C₁₈H₃₆                                            

molecular formula is C₁₈H₃₆  

8 0
3 years ago
The concentration of hydrogen peroxide solution can be determined by
max2010maxim [7]

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- =  0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of  MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since  

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution =  5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

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