The units of molar mass are?

A 15.00-ml sample of a naoh solution of unknown concentration requires 17.88 ml of a 0.1053 m h2so4 solution to reach the equiva

Sergeu [11.5K]

Answer:

<u>0.1255 M</u>

Explanation:

Base: NaOH

Vb = 15.00 ml = 15.00 / 1,000 liter

Mb = ?

Acid: H₂SO₄

Va = 17.88 ml = 17.88 / 1,000 liter

Ma = 0.1053

<u>2) Chemical reaction:</u>

The <em>titration</em> is an acid-base (neutralization) reaction to yield a salt and water:

Acid + Base → Salt + Water

H₂SO₄ (aq) + NaOH(aq) → Na₂SO₄ (aq) + H₂O (l)

<u>3) Balanced chemical equation:</u>

H₂SO₄ (aq) + 2 NaOH(aq) → Na₂SO₄ (aq) + 2H₂O (l)

Placing coefficient 2 in front of NaOH and H₂O balances the equation

<u>4) Stoichiometric mole ratio:</u>

The coefficients of the balanced chemical equation show that 1 mole of H₂SO₄ react with 2 moles of NaOH. Hence, the mole ratio is:

1 mole H₂SO₄ : 2 mole NaOH

<u>5) Calculations:</u>

a) Molarity formula: M = n / V (in liter)

⇒ n = M × V

b) Nunber of moles of acid:

nₐ = Ma × Va = 0.1053 (17.88 / 1,000)

c) Number of moles of base, nb:

nb = Mb × Vb = Mb × (15.00 / 1,000)

d) At equivalence point number of moles of acid = number of moles of base

nₐ = nb

0.1053 × (17.88 / 1,000) = Mb × (15.00 / 1,000)

Mb = 0.1053 × 17.88 / 15.00 = 0.1255 mole/liter = 0.1255 M

3 0

3 years ago

Read 2 more answers

in a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

dimulka [17.4K]

This is an incomplete question, here is a complete question.

Nitroglycerine (C₃H₅N₃O₉) explodes with tremendous force due to the numerous gaseous products. The equation for the explosion of Nitroglycerine is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

A scientist conducts an experiment to characterize a bomb containing nitroglycerine. She uses a steel, ridge container for the test.

Volume of rigid steel container: 1.00 L

Molar mass of Nitroglycerine: 227 g/mol

Temperature: 300 K

Amount of Nitroglycerine tested: 227 g

Value for ideal gas constant, R: 0.0821 L.atm/mol.K

In a second experiment, the total pressure is observed to be 58 atm. what is the partial pressure of the water vapor produced?

Answer : The partial pressure of the water vapor is, 20.01 atm

Explanation :

First we have to calculate the moles of $C_3H_5N_3O_9$

$\text{Moles of }C_3H_5N_3O_9=\frac{\text{Given mass }C_3H_5N_3O_9}{\text{Molar mass }C_3H_5N_3O_9}=\frac{227g}{227g/mol}=1mol$

Now we have to calculate the moles of $CO_2,O_2,N_2\text{ and }H_2O$

The balanced chemical reaction is:

$4C_3H_5N_3O_9(l)\rightarrow 12CO_2(g)+O_2(g)+6N_2(g)+10H_2O(g)$

From the balanced chemical reaction we conclude that,

As, 4 moles of $C_3H_5N_3O_9$ react to give 12 moles of $CO_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{12}{4}=3$ moles of $CO_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 1 moles of $O_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{1}{4}=0.25$ moles of $O_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 6 moles of $N_2$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{6}{4}=1.5$ moles of $N_2$

and,

As, 4 moles of $C_3H_5N_3O_9$ react to give 10 moles of $H_2O$

So, 1 moles of $C_3H_5N_3O_9$ react to give $\frac{10}{4}=2.5$ moles of $H_2O$

Now we have to calculate the mole fraction of water.

$\text{Mole fraction of }H_2O=\frac{\text{Moles of }H_2O}{\text{Moles of }H_2O+\text{Moles of }CO_2+\text{Moles of }O_2+\text{Moles of }N_2}$

$\text{Mole fraction of }H_2O=\frac{2.5}{2.5+3+0.25+1.5}=0.345$

Now we have to calculate the partial pressure of the water vapor.

According to the Raoult's law,

$p_{H_2O}=X_{H_2O}\times p_T$

where,

$p_{H_2O}$ = partial pressure of water vapor gas = ?

$p_T$ = total pressure of gas = 58 atm

$X_{H_2O}$ = mole fraction of water vapor gas = 0.345

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

$p_{H_2O}=X_{H_2O}\times p_T$

$p_{H_2O}=0.345\times 58atm=20.01atm$

Therefore, the partial pressure of the water vapor is, 20.01 atm

3 0

3 years ago

Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.

Vsevolod [243]

The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL

<h3>Balanced equation </h3>

2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

The mole ratio of the acid, H₃PO₄ (nA) = 2
The mole ratio of the base, Ca(OH)₂ (nB) = 3

<h3>How to determine the volume of Ca(OH)₂ </h3>

Molarity of acid, H₃PO₄ (Ma) = 0.390 M
Volume of acid, H₃PO₄ (Va) = 24.5 mL
Molarity of base, Ca(OH)₂ (Mb) = 0.279 M

Volume of base, Ca(OH)₂ (Vb) =?

MaVa / MbVb = nA / nB

(0.39 × 24.5) / (0.279 × Vb) = 2/3

9.555 / (0.279 × Vb) = 2/3

Cross multiply

2 × 0.279 × Vb = 9.555 × 3

0.558 × Vb = 28.665

Divide both side by 0.558

Vb = 28.665 / 0.558

Vb = 51.4 mL

Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL

Learn more about titration:

brainly.com/question/14356286

5 0

2 years ago

(WILL MARK BRAINLIEST)<br> PLZ HELP ASAP

ArbitrLikvidat [17]

Answer:

a) 2NaOH(aq) + CuSO4(aq) -------------> Cu(OH)2(s) + Na2SO4(aq)

b) Ca(OH)2(aq) + CO2(g) --------------> CaCO3 + H2O (this is already balanced)

c) Pb(NO3)2 + H2SO4 --------> PbSO4 + 2HNO3.

d) 2KNO3 ------> 2KNO2 + O2

e) H2SO4 + 2(NaOH) -----> Na2SO4 + 2(H2O)

f) Ca(NO3)2(aq) + (NH4)2CO3(aq) ----------------> CaCO3(s) + 2NH4NO3(aq)

3 0

4 years ago

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Calculate number of moles in 86.4 g of Ni

Ira Lisetskai [31]

Number of moles = mass of Ni /molecular mass of Ni

mass of nickel = 86.4 g
molecular mass of nickel = 58.69
number of moles of Ni in 86.4 g
=86.4/58.69
=1.472 mol
(rounded to four significant figures instead of three because the first digit of the answer starts with a 1).

4 0

3 years ago