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2 years ago

Calculate the molality of a 20.0% by mass ammonium sulfate (nh4)2so4 solution. the density of the solution is 1.117 g/ml.

1 answer:

7 0

Hello!

We have the following data:

m1 (solute mass) = 20 % m/m
M1 (Molar mass of solute) (NH4)2 SO4 = ?
m2 (mass of the solvent) = ? (in Kg)

First we find the solute mass (m1), knowing that:

20% m/m = 20g/100mL

20 ------ 100 mL (0,1 L)
y g --------------- 1 L

y = 20/0,1
y = 200 g --> m1 = 200 g

Let's find Solute's Molar Mass, let's see:

M1 of (Nh4)2SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

We must find the volume of the solvent and therefore its mass (m2), let us see:

d = 1,117 g/mL
m = 200 g
v (volumen of solute) = ?

$d = \dfrac{m}{V} \to V = \dfrac{m}{d}$

$V = \dfrac{200\:\diagup\!\!\!\!g}{1,117\:\diagup\!\!\!\!g/mL} \to V = 179\:mL\:(volumen\:of\:solute)$

The solvent volume will be:

1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

Then the mass of the solvent is:

m2 (mass of the solvent) = 821 g → m2 (mass of the solvent) = 0,821 Kg

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{m_1}{M_1*m_2}$

$\omega = \dfrac{200}{132*0,821}$

$\omega = \dfrac{200}{108,372}$

$\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

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Another way to find the answer:

We have the following data:

W (molality) = ? (in molal)
n (number of mols) = ?
m1 (solute mass) = 20 % m/m = 20g/100mL → (in g to 1L) = 200 g
m2 (disolvent mass) the remaining percentage, in the case: 80 % m/m = 800 g → m2 (disolvent mass) = 0,8 Kg
M1 (Molar mass of solute) (NH4)2 SO4
N = 2*14 = 28
H = (2*4)*1 = 8
S = 1*32 = 32
O = 4*16 = 64
----------------------
M1 of (Nh4)2SO4 = 28+8+32+64 => M1 = 132 g/mol

Let's find the number of mols (n), let's see:

 $n = \dfrac{m_1}{M_1}$

$n = \dfrac{200}{132}$

$n \approx 1,5\:mol$

Now, we apply all the data found to the formula of Molality, let us see:

$\omega = \dfrac{n}{m_2}$

$\omega = \dfrac{1,5}{0,8}$

$
\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark$

I hope this helps. =)

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A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

$AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

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

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V

4 0

2 years ago

Consider the solutions, 0.04 m urea [(NH2)2C=O)], 0.04 m AgNO3 and 0.04 m CaCl2. Which has (i) the highest osmotic pressure, (ii

Lana71 [14]

Answer:

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

Explanation:

The colligative properties depend upon the number of solute particles in a solution.

The following four are the colligative properties:

a) osmotic pressure : more the concentration of the solute, more the osmotic pressure

b) vapor pressure: more the concentration of the solute, lesser the vapor pressure.

c) elevation in boiling point: more the concentration of the solute, more the boiling point.

d) depression in freezing point: more the concentration of the solute, lesser the freezing point.

the number of particle produced by urea = 1

the number of particle produced by AgNO3 = 2

the number of particle produced by CaCl2 = 3

As concentrations are same, CaCl2 will have more number of solute particles and urea will have least

i) Highest osmotic pressure: CaCl2

ii) lower vapor pressure : CaCl2

iii) highest boiling point : CaCl2

5 0

2 years ago

On a clear day at sea level, with a temperature of 25 °C, the partial pressure of N2 in air is 0.78 atm and the concentration of

joja [24]

Answer : The partial pressure of nitrogen gas is, 2.94 atm

Explanation:

According top the Henry's Law, the concentration of a gas in a liquid is directly proportional to the partial pressure of the gas.

$C\propto P$

$C=K_H\times P$

$K_H$ is Henry's constant.

or,

$\frac{C_1}{C_2}=\frac{P_1}{P_2}$

where,

$C_1$ = initial concentration of gas = $5.3\times 10^{-4}M$

$C_2$ = final concentration of gas = $2.0\times 10^{-3}M$

$P_1$ = initial partial pressure of gas = 0.78 atm

$P_2$ = final partial pressure of gas = ?

Now put all the given values in the above formula, we get the final partial pressure of the gas.

$\frac{5.3\times 10^{-4}M}{2.0\times 10^{-3}M}=\frac{0.78atm}{P_2}$

$P_2=2.94atm$

Therefore, the partial pressure of nitrogen gas is, 2.94 atm

5 0

2 years ago

Each day, the stomach produces 2.0 L of gastric juice that contains 0.10 M HCl. Phillips Milk of Magnesia is a white-colored, aq

trasher [3.6K]

Answer:

It would take 72.9 mL of milk of magnesia.

Explanation:

First of all we have to think how the compounds react with each other and what are the products formed. In this case, the hydrochloric acid reacts with magnesium hydroxide to generate magnesium chloride and water as a subproduct. Having said that, we have to state the balanced chemical reaction to know the associated stoichiometry:

2 HCl + Mg(OH)2 → MgCl2 + 2 H2O

According to the balanced equation we know that 2 mol of HCl reacts with 1 mol of Mg(OH)2.

Now we calculate the quantity of moles of HCl that we have present in 2.0 lts of 0.10 M solution:

0.1 M HCl = 0.1 moles HCl / 1000 ml Solution

So, in 2 liters of solution we will have 0.2 moles of HCl

This 0.2 moles of acid, as we stated before, will react with 0.1 moles of Mg(OH)2, so we need to calculate the amount of milk of magnesia that has this required quantity of moles.

With the molar mass of Mg(OH)2 we calculate the weight of the compound that represents the 0.1 moles needed to react with all the HCl present in solution:

1 mol Mg(OH)2 = 58.32 g

0.1 mol = 5.832 g

Now we need to determine what volume of the milk of magnesia solution has 5.832 g of Mg(OH)2 to react with the acid:

The concentration of milk of magnesia is 8 % (w/v). This means that we have 8 gr of Mg(OH)2 per 100 ml of solution.

8 gr Mg(OH)2 per 100 mL Solution

5.832 gr Mg(OH)2 = 72.9 mL of Milk of Magnesia

6 0

2 years ago

Be sure to answer all parts. Consider the following balanced redox reaction (do not include state of matter in your answers): 2C

timurjin [86]

Answer:

The specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

Explanation:

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2CrO_2^- + 6ClO^- + 2H_2O\rightarrow 2CrO_4^{2-} + 3Cl_2 + 4OH^-$

The half cell reactions for the above reaction follows:

Oxidation half reaction: $CrO_2^- + 2H_2O + 4OH^-\rightarrow CrO_4^{2-} + 4H_2O + 3e^-$

Reduction half reaction: $2ClO^- + 4H_2O + 2e^-\rightarrow Cl_2 + 2H_2O + 4OH^-$

Thus, the specie which is oxidized is:- $CrO_2^-$

The specie which is reduced is:- $ClO^-$

8 0

3 years ago