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Snowcat [4.5K]
3 years ago
6

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

ution?
Chemistry
1 answer:
jeka57 [31]3 years ago
4 0

<u>Answer:</u> The molality of non-electrolyte is 24.69 m

<u>Explanation:</u>

We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g

To calculate the molality of solution, we use the equation:

\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}

Where,

n_{solute} = Moles of solute (non-electrolyte) = 0.310 moles

W_{solvent} = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m

Hence, the molality of non-electrolyte is 24.69 m

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What is the pH of 0.26 M ammonium ion? NH4+(aq) + H2O(1) NH3(aq) + H30* (aq) a. 4.33 b.9.25 c. 3.87 d. 4.92 e. 4.75
choli [55]

Answer:

b) pH = 9.25

Explanation:

  • NH4+(aq)  +  H2O(l)  ↔  NH3(aq)  +  H3O+(aq)
  • NH3 + H2O ↔ NH4+  +  OH-
  • 2 H2O ↔ H3O+  +  OH-

⇒ Kb = [ NH4+ ] * [ OH- ] / [ NH3 ] = 1.86 E-5......from literature

mass balance NH4+:

⇒ M NH4+ = [ NH4+ ] - [ OH- ]

∴ [ NH3 ] ≅ M NH4+ = 0.26 M

⇒ Kb = (( 0.26 + [ OH- ] )) * [ OH- ] / 0.26 = 1.86 E-5

⇒ 0.26 [ OH-] + [ OH- ]² = 4.836 E-6

⇒ [ OH- ]² + 0.26 [ OH- ] - 4.836 E-6 = 0

⇒ [ OH- ] = 1.859 E-5 M

⇒ pOH = - Log ( 1.859 E-5 )

⇒ pOH = 4.7305

⇒ pH = 14 - pOH = 9.269

6 0
3 years ago
A system fitted with a piston expands when it absorbs 53.1 ) of heat from the surroundings. The piston is working against a pres
Hoochie [10]

Answer:

d. 60.8 L

Explanation:

Step 1: Given data

  • Heat absorbed (Q): 53.1 J
  • External pressure (P): 0.677 atm
  • Final volume (V2): 63.2 L
  • Change in the internal energy (ΔU): -108.3 J

Step 2: Calculate the work (W) done by the system

We will use the following expression.

ΔU = Q + W

W = ΔU - Q

W = -108.3 J - 53.1 J = -161.4 J

Step 3: Convert W to atm.L

We will use the conversion factor 1 atm.L = 101.325 J.

-161.4 J ×  1 atm.L/101.325 J = -1.593 atm.L

Step 4: Calculate the initial volume

First, we will use the following expression.

W = - P × ΔV

ΔV = - W / P

ΔV = - 1.593 atm.L / 0.677 atm = 2.35 L

The initial volume is:

V2 = V1 + ΔV

V1 = V2 - ΔV

V1 = 63.2 L - 2.35 L = 60.8 L

7 0
3 years ago
What is the atomic mass equal to?
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Atomic mass is equal to the total number of electrons  neutrons and protons

8 0
3 years ago
Read 2 more answers
Determine the [H3O+] in a 0.265 M HClO solution. The Ka of HClO is 2.9 × 10-8.
dexar [7]

8.8 × 10-5 M is the  [H3O+] concentration in 0.265 M HClO solution.

Explanation:

HClO is a weak acid and does not completely dissociate in water as ions.

the equation of dissociation can be written and ice table to be formed.

 HClO +H2O ⇒ ClO- + H3O+

I  0.265                0        0

C  -x                    +x     +x

E  0.265-x          +x      +x

Now applying the equation of Ka, where Ka is given as 2.9 × 10-8.

Ka = \frac{[ClO-][H3O+]}{[HClO]}

2.9 × 10^-8 = \frac{[x] [x]}{[0.265-x]}

x^{2} = 7.698 x10^{-9}

x = 8.8 × 10-5 M

The hydronium ion concentration is 8.8 × 10-5 M  in 0.265 M solution of HClO.

8 0
3 years ago
. How many milliliters of 0.20 M HCl are needed to exactly neutralize 40. milliliters of 0.40 M KOH
Anuta_ua [19.1K]

Answer:

V_{HCl}=80mL

Explanation:

Hello,

In this case, for the given reactants we identify the following chemical reaction:

KOH+HCl\rightarrow KCl+H_2O

Thus, we evidence a 1:1 molar ratio between KOH and HCl, therefore, for the complete neutralization we have equal number of moles, that in terms of molarities and volumes become:

n_{HCl}=n_{KOH}\\\\M_{HCl}V_{HCl}=M_{KOH}V_{KOH}

Hence, we compute the volume of HCl as shown below:

V_{HCl}=\frac{M_{KOH}V_{KOH}}{M_{HCl}} =\frac{0.40M*40mL}{0.20M} \\\\V_{HCl}=80mL

Best regards.

5 0
3 years ago
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