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diamong [38]
3 years ago
14

Acetonitrile (CH3CN) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a 1

.80 M LiBr solution in acetonitrile is 0.824 g/cm^3 (cubed)
a.) Calculate the concentration of the solution in molality.
b.) Calculate the concentration of the solution in mole fraction of LiBr
c.) Calculate the concentration of the solution in mass percentage of CH3CN
Chemistry
1 answer:
erastovalidia [21]3 years ago
3 0

Answer:

a. [LiBr] = 2.70 m

b. Xm for LiBr = 0.1

c. 81% by mass CH₃CN

Explanation:

Solvent → Acetonitrile (CH₃CN)

Solute → LiBr, lithium bromide

We convert the moles of solute to mass → 1.80 mol . 86.84 g/1 mol = 156.3 g

This mass of solute is contained in 1L of solution

1 L = 1000 mL → 1mL = 1cm³

We determine solution mass by density

Solution density = Solution mass / Solution volume

Solution density . Solution volume = solution mass

0.824 g/cm³ . 1000 cm³ = 824 g

Mass of solution = 824 g (solvent + solute)

Mass of solute = 156.3 g

Mass of solvent = 824 g - 156.3 g = 667.7 g

Molality → Moles of solute in 1kg of solvent

We convert the mass of solvent from g to kg → 667.7 g . 1kg /1000g = 0.667 kg

Mol/kg → 1.80 mol / 0.667 kg = 2.70 m → molality

Mole fraction → Mole of solute / Total moles (moles solute + moles solvent)

Moles of solvent → 667.7 g . 1mol/ 41g = 16.3 moles

Total moles = 16.3 + 1.8 = 18.1

Mole fraction Li Br → 1.80 moles / 18.1 moles = 0.1

Mass percentage → (Mass of solvent, <u>in this case</u> / Total mass) . 100

<u>We were asked for the acetonitrile</u> → (667.7 g / 824 g) . 100 = 81%

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Read 2 more answers
15 pts The drop down items have the same options for all sections labeled choose an item
tia_tia [17]

Answer:

#8 : 21.5 L of CH₄

#9 : 36 g of H₂O

#9 : 5.4 x 10²² molecules of CO₂

Explanation:

#8 : Part 1.

Data given:

no. of moles of CH₄ = 0.960 mol

volume of CH₄ = ?

Solution

volume can be calculated by following formula

          No. of moles = Volume of gas / molar volume

Rearrange the above equation

            Volume of gas = No. of moles x molar volume . . . . . . (1)

Where

molar volume = 22.4 L/mol

Put values in equation 1

           Volume of gas = 0.960 mol x 22.4 L/mol

           Volume of gas = 21.5 L

So,

By calculation 0.960 moles have 21.5 L volume of CH₄

_______________

#9 : Part 2.

Data given:

no. of moles of H₂O = 2.0 mol

mass of H₂O = ?

Solution

volume can be calculated by following formula

          No. of moles = mass in grams / molar mass

Rearrange the above equation

            mass in grams = No. of moles x molar mass . . . . . . (2)

Where

molar mass of H₂O = 2 (1) + 16

molar mass of H₂O = 18 g/mol

Put values in equation 2

      mass in grams = 2.0 mol x 18 g/mol

        mass in grams = 36 g

So,

By calculation 2.0 moles have 36 g mass of H₂O

________________

#10 : Part 3.

Data given:

volume of CO₂ = 2 L

no. of molecules of CO₂ = ?

Solution

First we have to find out number of moles of CO₂

Following formula will be used

          No. of moles = Volume of gas / molar volume

Where

molar volume = 22.4 L/mol

Put values in above equation

           No. of moles = 2 L / 22.4 L/mol

           No. of moles = 0.0893 mol

So,

No. of moles of CO₂ = 0.0893 mol

Now

we will calculate number of molecules by using following formula

          No. of moles = no. of molecules / Avogadro's number

Rearrange the above equation

            no. of molecules = No. of moles x Avogadro's number . . . . . . (3)

Where

Avogadro's number = 6.022 x 10²³

Put values in above equation 3

            no. of molecules =  0.0893 mol x 6.022 x 10²³

            no. of molecules =  5.4 x 10²²

So,

By calculation 2 L of CO₂ have 5.4 x 10²² molecules of CO₂

8 0
4 years ago
Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.
TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

HCl+KOH\rightarrow KCl+H_2O

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

n_{KOH}=0.00768mol-0.00192mol=0.00576mol

And the resulting concentration of KOH and OH ions as this is a strong base:

[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M

And the resulting pH is:

pH=14+log(0.131)\\\\pH=13.1

Regards!

3 0
3 years ago
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