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Karolina [17]
2 years ago
6

You have a 150mL sample of an aqueous solution at 25C. It contains 15.2mg of an unknown nonelectrolyte compound. If the solution

has an osmotic pressure of 8.44 torr, what is the molar mass of the compound
Chemistry
1 answer:
Serga [27]2 years ago
7 0

The molar mass of the compound:

If the solution has an osmotic pressure of 8.44 torr, then the molar mass of the unknown non-electrolyte is 223.14 g.

What is osmosis?

  • Osmosis is defined as the flow of solvent molecules through semi-permeable membrane.
  • Osmotic pressure is the pressure applied to stop the flow of solvent molecules.
  • It is a colligative property that means osmotic pressure depends on the number of solute particles .

Therefore,

πV=inRT  ( for electrolytes)

Where,   π= Osmotic pressure

i = Van 't Hoff factor

n= moles

R= Gaseous constant = 62.363577 L torr mol^{-1}K^{-1}

T= Temperature

V= Volume of solution

Given:

T= 298K

V= 150 mL= 0.150 L

Given mass of unknown electrolyte= 15.2 mg = 15.2 x  10^{-3} g

Osmotic pressure= 8.44 torr

Molar mass= ?

For non-electrolytes:

πV = n RT

πV=\frac{m}{M}RT

Calculations:

Putting the given values in the formula:

8.44 x 0.150 =15.2 x 10^{-3}/ M x 62.36 x 298

1.266 = 282.5/M

M = 282.5/1.266

M = 223.14 g

Therefore,

The molar mass of the unknown non-electrolyte is 223.14g.

Learn more about Osmotic pressure here,

brainly.com/question/13680877

#SPJ4

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mr_godi [17]

A is the answer

In an ozone molecule, the three atoms must be connected, so there must at least be a single bond between them. Place

dots in pairs around the oxygen atoms until each oxygen atom has eight valence electrons, starting with the atoms on the

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identified in part A. Remember that the dashes between the oxygen atoms, which represent single bonds, each indicate

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6 0
3 years ago
Plz do 108, 110, and 112! I really need help
Gennadij [26K]
Im sorry i can only do the last one the answer might be a, d, e. Im so sorry if im wrong
8 0
3 years ago
Read 2 more answers
The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57
In-s [12.5K]

Answer:

1.135 M.

Explanation:

  • For the reaction: <em>2HI → H₂ + I₂,</em>

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

  • To solve this problem, we can use the integral law of second-order reactions:

<em>1/[A] = kt + 1/[A₀],</em>

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>

8 0
3 years ago
Carbon dioxide dissolves in water to form carbonic acid, which is primarily dissolved CO2. Dissolved CO2 satisfies the equilibri
lorasvet [3.4K]

Explanation:

The reaction equation will be as follows.

           CO_{2}(aq) + H_{2}O \rightleftharpoons H^{+}(aq) + HCO^{-}_{3}(aq)

Calculate the amount of CO_{2} dissolved as follows.

             CO_{2}(aq) = K_{CO_{2}} \times P_{CO_{2}}

It is given that K_{CO_{2}} = 0.032 M/atm and P_{CO_{2}} = 1.9 \times 10^{-4} atm.

Hence, [CO_{2}] will be calculated as follows.

           [CO_{2}] = K_{CO_{2}} \times P_{CO_{2}}          

                           = 0.032 M/atm \times 1.9 \times 10^{-4}atm

                           = 0.0608 \times 10^{-4}

or,                        = 0.608 \times 10^{-5}

It is given that K_{a} = 4.46 \times 10^{-7}

As,      K_{a} = \frac{[H^{+}]^{2}}{[CO_{2}]}

          4.46 \times 10^{-7} = \frac{[H^{+}]^{2}}{0.608 \times 10^{-5}}  

               [H^{+}]^{2} = 2.71 \times 10^{-12}

                      [H^{+}] = 1.64 \times 10^{-6}

Since, we know that pH = -log [H^{+}]

So,                      pH = -log (1.64 \times 10^{-6})

                                 = 5.7

Therefore, we can conclude that pH of water in equilibrium with the atmosphere is 5.7.

3 0
3 years ago
Which type of element seeks to lose electrons
kvv77 [185]
Metals because, they have less than 3 electrons in the outermost shell and they usually lose electrons.
6 0
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