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butalik [34]
3 years ago
15

What is the molarity of each ion in a solution prepared by dissolving 0.520 g of na2so4, 1.186 g of na3po4, and 0.223 g of li2so

4 in water and diluting to a volume of 100.00 ml?
Chemistry
1 answer:
Leto [7]3 years ago
8 0
First  calculate  the  number  of moles  of  each  ion  present
for  Na2So4  is
0.52/142=0.0037moles
Na+  =0.0037  x2 =0.0074moles
SO4^-2  =0.0027moes

for Na3PO4=  1.186/164=0.0072 moles
Na+   =0.0072  x3=0.0216moles
po4  ions =  0.0072  moles

for LiSo4 =0.223/110=0.002moles
Li  ions=0.002 x2  =0.004mole
SO4  ions =0.002moles
total moles  for
Na  ions  =0.0216  +0.0074 =0.029moles
forSO4 ions =0.0037 +0.002=0.0057moles
molarity is  therefore
Na  ion=0.029/0.1=0.29M
SO4  ions=0.0057/0.1=0.057M
Li  ions =0.004/0.1=0.04M
PO4  ions=0.0072/0.1=0.072M

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In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

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\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}

It means we can compute the final density as shown below:

\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}

Now, we plug in to obtain:

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