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nevsk [136]
4 years ago
13

100 g of an inorganic compound X.5H2O containing a volatile impurity was kept in an oven at 150 degree celcius for 60 minutes. t

he weight of residue after heating is 8g. the percentage of impurity in X was?
Chemistry
1 answer:
Hoochie [10]4 years ago
6 0
              150°C
X*5H₂O    →    X + 5H₂O↑

m₀=100 g
m(X)=8 g
M(H₂O)=18.0 g/mol
n(H₂O)=5 mol

the mass of the water 
m(H₂O)=M(H₂O)n(H₂O)
m(H₂O)=18.0*5=90 g

the mass of the residue after removal of water
m'(X)=100-90=10 g

the percentage of impurity in X*5H₂O
w(imp)=[m'(X)-m(X)]/m₀

w(imp)=[10-8]/100=0.02 (2%)
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<h3>Answer:</h3>

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<h3>Explanation:</h3>

Concept tested: Moles and Molarity

In this case we are give;

Mass of solid sodium hydroxide as 13.20 g

Molarity of H₂SO₄ as 0.235 M

We are required to determine the volume of H₂SO₄ required

<h3>First: We need to write the balanced equation for the reaction.</h3>
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<h3>Second: We calculate the umber of moles of NaOH used </h3>
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<h3>Third: Determine the number of moles of the acid, H₂SO₄</h3>
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  • Therefore, the mole ratio of NaOH: H₂SO₄ is 2 : 1.
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                                    = 0.33 moles × 2

                                   = 0.66 moles of H₂SO₄

<h3>Fourth: Determine the Volume of the acid, H₂SO₄ used</h3>
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In this case;

Volume of the acid = 0.66 moles ÷ 0.235 M

                                = 2.809 L

Therefore, the volume of the acid required to neutralize the base,NaOH is 2.809 L.

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