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2 years ago

Calculate the molarity of 48.0 mL of 6.00 M H2SO4 diluted to 0.250 L

Chemistry

2 answers:

const2013 [10]2 years ago

4 0

Answer:

The answer is 1.15m.

Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.

We can find the number of H2SO4 moles by using its molarity

C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288

Since water has a density of 1.00kgL, the mass of solvent is

m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg

Therefore, molality is

m=nmass.solvent=0.288moles0.250kg=1.15m

shusha [124]2 years ago

3 0

Answer:

The molarity of 48.0 mL of 6.00 M H₂SO₄ diluted to 0.250 L is 1.152 M

Explanation:

Dilution is the procedure used to prepare a less concentrated solution from a more concentrated solution. It consists of adding solvent to an existing solution. Then the amount of solute does not vary, but the volume of the solvent does: when more solvent is added, the concentration of the solute decreases, as the volume of the solution increases.

When working with dilutions, the following can be applied, taking into account that the mass of the solute does not change and neither does its moles:

moles of solute before dilution = moles of solute after dilution

Mi*Vi=Mf*Vf

where

Mi: initial molarity of the solution.
Vi: initial volume of the solution.
Mf: final molarity of the solution.
Vf: final volume of the solution.

In this case, the following values are possessed:

Mi= 6 M
Vi= 48 mL=0.048 L (1 L=1000 mL)
Mf= ?
Vf= 0.250 L

Replacing:

6 M*0.048 L=Mf* 0.250 L

Resolving:

$Mf=\frac{6M*0.048L}{0.250 L}$

Mf=1.152 M

The molarity of 48.0 mL of 6.00 M H₂SO₄ diluted to 0.250 L is 1.152 M

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Lithium acetate, LiCH3CO2, is a salt formed from the neutralization of the weak acid acetic acid, CH3CO2H, with the strong base

Vesna [10]

Answer : The pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

Explanation :

First we have to calculate the value of $K_b$ .

As we know that,

$K_a\times K_b=K_w$

where,

$K_a$ = dissociation constant of an acid = $1.8\times 10^{-5}$

$K_b$ = dissociation constant of a base = ?

$K_w$ = dissociation constant of water = $1\times 10^{-14}$

Now put all the given values in the above expression, we get the dissociation constant of a base.

$1.8\times 10^{-5}\times K_b=1\times 10^{-14}$

$K_b=5.5\times 10^{-10}$

Now we have to calculate the concentration of hydroxide ion.

Formula used :

$[OH^-]=(K_b\times C)^{\frac{1}{2}}$

where,

C is the concentration of solution.

Now put all the given values in this formula, we get:

$[OH^-]=(5.5\times 10^{-10}\times 0.289)^{\frac{1}{2}}$

$[OH^-]=1.3\times 10^{-5}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.3\times 10^{-5})$

$pOH=4.9$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-4.9=9.1$

Therefore, the pH of 0.289 M solution of lithium acetate at $25^oC$ is 9.1

4 0

2 years ago

A 500 mL sample of a 0.100 M formate buffer, pH 3.75, is treated with 5 mL of 1.00 M KOH. What is the pH following this addition

lubasha [3.4K]

Answer: The pH of the solution after addition of KOH is 3.84

Explanation:

We are given:

pH of buffer = 3.75

$pK_a$ of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

$pH=pK_a+\log(\frac{[HCOO-]}{[HCOOH]})$

Putting values in above equation, we get:

$3.75=3.75+\log(\frac{[HCOO-]}{[HCOOH]})\\\\\log(\frac{[HCOO-]}{[HCOOH]})=0\\\\\frac{[HCOO-]}{[HCOOH]}=1$

$[HCOO-]=[HCOOH]$

We are given:

Concentration of formate buffer = 0.100 M

$[HCOO-]+[HCOOH]=0.1$

$[HCOO-]=[HCOOH]=0.05M$

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

$1.00M=\frac{\text{Moles of KOH}\times 1000}{5mL}\\\\\text{Moles of KOH}=0.005mol$

The chemical reaction for formic acid and KOH follows the equation:

$HCOOH+KOH\rightarrow HCOO^-+H_2O$

Initial: 0.05 0.005 0.05

Final: 0.045 - 0.055

Volume of solution = 500 + 5 = 505 mL = 0.505 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[HCOO^-]}{[HCOOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of formic acid = 3.75

$[HCOO^-]=\frac{0.055}{0.505}$

$[HCOOH]=\frac{0.045}{0.505}$

pH = ?

Putting values in above equation, we get:

$pH=3.75+\log(\frac{0.055/0.505}{0.045/0.505})\\\\pH=3.84$

Hence, the pH of the solution after addition of KOH is 3.84

3 0

2 years ago

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

Flauer [41]

Step 1: Change density from g/mL to g/L;

0.807 g/mL = 807 g/L

Step 2: Find Moles of N₂;
As,
Density = Mass / Volume
Or,
Mass = Density × Volume

Putting Values,

Mass = 807 g/L × 1 L

Mass = 807 g
Also,
Moles = Mass / M.mass

Putting values,

Moles = 807 g / 28 g.mol⁻¹

Moles = 28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
P V = n R T

V = n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

V = (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

V = 704.76 L

8 0

2 years ago

Compound A is an organic compound which contains Carbon, Hydrogen and Oxygen. When 0.240g of the vapour of A is slowly passed ov

Keith_Richards [23]

Answer:

1) 0.009 61 g C; 2) 0.008 00 mol C

Step-by-step explanation:

You know that you will need a balanced equation with masses, moles, and molar masses, so gather all the information in one place.

M_r: 12.01 44.01

C + ½O₂ ⟶ CO₂

m/g: 0.352

1) Mass of C

Convert grams of CO₂ to grams of C

44.01 g CO₂ = 12.01 g C

Mass of C = 0.352 g CO₂ × 12.01 g C/44.01 g CO₂

Mass of C = 0.009 61 g C

2) Moles of C

Convert mass of C to moles of C.

1 mol C = 12.01 g C

Moles of C = 0.00961 g C × (1 mol C/12.01 g C)

Moles of C = 0.008 00 mol C

All the carbon comes from Compound A, so there are 0.008 00 mol C in Compound A.

7 0

2 years ago

How many milliliters of 0.258M NaOH are required to completely neutralize 2.00 g of acetic acid HC2H3O2?

madam [21]

Answer:

0.129 L = 129.0 mL.

Explanation:

NaOH neutralizes acetic acid (CH₃COOH) according to the balanced reaction:

NaOH + CH₃COOH → CH₃COONa + H₂O.

According to the balanced equation: 1.0 mole of NaOH will neutralize 1.0 mole of CH₃COOH.

no. of moles of CH₃COOH = mass/molar mass = (2.0 g)/(60 g/mol) = 0.033 mole.

For NaOH:

no. of moles = (0.258 mol/L)(V)

At neutralization: no. of moles of NaOH = no. of moles of CH₃COOH

∴ (0.258 mol/L)(V) = 0.033 mole

∴ The volume of NaOH = (0.033 mole)/(0.258 mol/L) = 0.129 L = 129.0 mL.

8 0

2 years ago