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3 years ago

Find molarity of 29.1 g LiClO4 · 3 H2O in 235 mL of solution

Chemistry

1 answer:

juin [17]3 years ago

3 0

To find molarity

We need no.of moles and solution in liter

Given data is in grams and ml

We have to convert it in required units

For no.of moles use formula

No.of mole=mass in gram / molar mass

Mass in gram = 29.1g

Molar mass = 160.4374g/mol calculated by adding the atomic wight of all atoms in LiClO4.3H2O

Atomic weight

Li= 6.941

Cl= 35.453

O= 15.9994

H= 1.00794

Now put it in mol equation

No.of mole = 29.1/160.4374

= 0.181mol of solute

Now convert solution in L from ml

235ml = 235/1000 = 0.235L

Now put these values in molarity formula

Molarity= 0.181/0.235

= 0.77M solution

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To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

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Volume of solution = 0.03340 L

Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol$

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$2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

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Excess moles of sulfuric acid = 0.00835 moles

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Putting values in equation 1, we get:

$0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol$

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$2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2$

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = $\frac{2}{3}\times 0.0556=0.0371mol$ of metal

To calculate the molar mass of metal for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

$0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol$

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Explanation:

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