Use the periodic table to help you.
2 answers:
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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) = ?
<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:
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<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>
Now, we apply all the data found to the formula of Molality, let us see:
I hope this helps. =)
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) = ?
<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:
_________________________________
_________________________________
<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>
Now, we apply all the data found to the formula of Molality, let us see:
I hope this helps. =)