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zvonat [6]
3 years ago
5

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

Chemistry
1 answer:
olasank [31]3 years ago
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)

<span>The solvent volume will be:
</span>
1000 -179 => V = 821 mL (volumen of disolvent)

If: 1 mL = 1g

<span>Then the mass of the solvent is:
</span>
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

_________________________________
_________________________________


<span>Another way to find the answer:
</span>
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 


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

</span>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}

&#10;\boxed{\boxed{\omega \approx 1,8\:Molal}}\end{array}}\qquad\checkmark

I hope this helps. =)
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