How many molecules of water are there in 8.050 x 10^3 grams of water

By titration, 15.0 mLmL of 0.1008 MM sodium hydroxide is needed to neutralize a 0.2053-gg sample of an organic acid. What is the

andre [41]

The question is incomplete, here is the complete question:

By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monoprotic.

Answer: The molar mass of monoprotic acid is 135.96 g/mol

Explanation:

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

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

For sodium hydroxide:

Molarity of NaOH solution = 0.1008 M

Volume of solution = 15.0 mL

Putting values in above equation, we get:

$0.1008M=\frac{\text{Moles of NaOH}\times 1000}{15.0}\\\\\text{Moles of NaOH}=\frac{(0.1008\times 15.0)}{1000}=0.00151mol$

As, the acid is monoprotic, it contains 1 hydrogen ion

1 mole of $OH^-$ ion of NaOH neutralizes 1 mole of $H^+$ ion of monoprotic acid

So, 0.00151 moles of $OH^-$ ion of NaOH will neutralize $\frac{1}{1}\times 0.00151=0.00151mol$ of $H^+$ ion of monoprotic acid

In monoprotic acid:

1 mole of $H^+$ ion is released by 1 mole of monoprotic acid

So, 0.00151 moles of $H^+$ ion will be released by $\frac{1}{1}\times 0.00151=0.00151mol$ of monoprotic acid

To calculate the number of moles, we use the equation:

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

Moles of monoprotic acid = 0.00151 mole

Given mass of monoprotic acid = 0.2053 g

Putting values in above equation, we get:

$0.00151mol=\frac{0.2053g}{\text{Molar mass of monoprotic acid}}\\\\\text{Molar mass of monoprotic acid}=\frac{0.2053g}{0.00151mol}=135.96g/mol$

Hence, the molar mass of monoprotic acid is 135.96 g/mol

8 0

2 years ago

What element is used in making thermometers

slava [35]

If it's the liquid inside:

mercury

3 0

3 years ago

Read 2 more answers

You are asked to prepare 500. mL of a 0.200 M acetate buffer at pH 5.10 using only pure acetic acid (MW=60.05 g/mol, pKa=4.76),

Alenkasestr [34]

Answer:

1. 6.005 g

2. 22.9 mL

3. Until the mixtures becomes homogeneous.

Explanation:

A buffer is a solution where a weak acid is in equilibrium with its conjugate base (its anion) or a weak base is in equilibrium with its conjugate base (its cation). The buffer remains the pH almost unaltered because it shifts the equilibrium if an acid or base is added.

1. The pH of a buffer can be calculated by the Henderson-Hasselbalch equation:

pH = pKa + log[A⁻]/[HA]

Where [A⁻] is the concentration of the conjugate base (the anion) of the acid, and HA is the acid concentration.

5.10 = 4.76 + log[A⁻]/[HA]

log[A⁻]/[HA] = 5.10 - 4.76

log[A⁻]/[HA] = 0.34

[A⁻]/[HA] = $10^{0.34}$

[A⁻]/[HA] = 2.1878

Because the volume is the same, we can replace the concentration by the number of moles (n):

nA⁻/nHA = 2.1878

nA⁻ = 2.1878*nHA

The total number of moles of the substances in the buffer is: 0.200 mol/L * 0.5 L = 0.1 mol

nA⁻ + nHA = 0.1

2.1878*nHA + n HA = 0.1

3.1878nHA = 0.1

nHA = 0.0314 mol

nA⁻ = 0.0686 mol

The total number of moles of acetic acid needed is 0.1 mol (both substances may be from it):

m = MW*mol

m = 60.05*0.1 = 6.005 g

2. NaOH must react with acetic acid to form the anion, so for a 1:1 reaction, it will be needed 0.0686 mol of NaOH:

V = mol/concentration

V = 0.0686/3

V = 0.0229 L = 22.9 mL

3. The buffer must be a homogeneous solution, it means that it can't be noticed phases in the buffer, so the flask must be inverted until all the buffer is diluted in water, and it will be noticed when the solution becomes homogenous.

6 0

3 years ago

What is the molality of an aqueous KCl solution with a mole fraction of KCl, XKCl = 0.175? (The molar mass of KCl = 74.55 g/mol

NARA [144]

Answer:

The molality of the KCl solution is 11.8 molal

Explanation:

Step 1: Data given

Mol fracrion KCl = 0.175

Molar mass KCl = 74.55 g/mol

Molar mass H2O = 18.02 g/mol

Step 2: Calculate mol fraction H2O

mol fraction H2O = 1 - 0.175 = 0.825

Step 3: Calulate mass of H2O

Suppose the total moles = 1.0 mol

Mass H2O = moles H2O * molar mass

Mass H2O = 0.825 * 18.02 g/mol

Mass H2O = 14.87 grams = 0.01487 kg

Step 4: Calculate molality

Molality KCl = 0.175 / 0.01487 kg

Molality KCl = 11.8 molal

The molality of the KCl solution is 11.8 molal

5 0

3 years ago

I need answers for this question.

Svetach [21]

I cant see the question im sorru

4 0

3 years ago