Respuesta :

dexteright02
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) = ?

[tex]d = \dfrac{m}{V} \to V = \dfrac{m}{d} [/tex]

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

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:

[tex]\omega = \dfrac{m_1}{M_1*m_2} [/tex]

[tex]\omega = \dfrac{200}{132*0,821} [/tex]

[tex]\omega = \dfrac{200}{108,372} [/tex]

[tex]\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark[/tex]

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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:

[tex]n = \dfrac{m_1}{M_1} [/tex]

[tex]n = \dfrac{200}{132}[/tex]

[tex]n \approx 1,5\:mol[/tex]

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

[tex]\omega = \dfrac{n}{m_2} [/tex]

[tex]\omega = \dfrac{1,5}{0,8} [/tex]

[tex] \boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark[/tex]

I hope this helps. =)

The molality is 1.88m.

Given:

Mass of (NH₄)₂SO₄=20.0%

Density=1.117g/mL

To find:

Molality=?

Step 1: Calculate mass of  (NH₄)₂SO₄:

Solute mass: 20 % m/m = 20g/100mL → (in g to 1L) = 200 g

Solvent mass: The remaining percentage, in the case: 80 % m/m = 800 g

Step 2: Calculate number of moles:

Molar mass of (NH₄)₂SO₄= 132.14g/mol

[tex]\text{Number of moles}=\frac{200}{132.14} =1.51mol[/tex]

Step 3: Calculation of molality:

∵(1kg=100g)

Molality is a measure of the number of moles of solute in a solution corresponding to 1 kg or 1000 g of solvent.

[tex]\text{Molality}=\frac{1.51 mol}{0.8 kg} =1.88m[/tex]

Learn more:

brainly.com/question/14846702

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