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3 years ago

What is the molarity of a solution containing 3.6 mol of KCl in 750 mL?

Chemistry

1 answer:

Vesnalui [34]3 years ago

4 0

Answer:

$\boxed {\boxed {\sf 4.8 \ M \ KCl}}$

Explanation:

The molarity of a solution is found by dividing the moles of solute by the liters. of solvent.

$molarity=\frac{moles}{liters}$

We know the solution has 3.6 moles of potassium chlorine. We know there are 750 milliliters of solvent, but we need to convert this to liters.

1 liter is equal to 1000 milliliters. Set up a proportion.

$\frac {1 \ L }{1000 \ mL}$

Multiply by 750 mL and the units of mL will cancel.

$750 \ mL * \frac {1 \ L }{1000 \ mL} = \frac {750 \ mL}{1000 \ mL}= 0.75 \ L$

Now we know the moles and liters, so we can calculate molarity.

moles= 3.6 mol KCl
liters= 0.75 L

Substitute these values into the formula.

$molarity= \frac{3.6 \ mol \ KCl}{0.75 \ L}$

Divide.

$molarity= 4.8 \ mol \ KCl/ L$

1 mole per liter is equal to 1 molar.
Our answer of 4.8 moles of potassium chloride per liter is equal to 4.8 M KCl

$molarity= 4.8 \ M \ KCl$

The molarity of the solution is 4.8 M KCl.

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Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

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Whereas the half reactions are:

$Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O$

Next, we exchange the transferred electrons:

$1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow 4N^{4+}O^{2-}_2+4H_2O$

Afterwards, we add them to obtain:

$Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O$

By adding and subtracting common terms we obtain:

$Sn^0+4H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O$

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$Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O$

Therefore, the smallest whole-number coefficient for Sn is 1.

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