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3 years ago

If 10.7 grams of NH4Cl is dissolved in enough water to make 800 mL of solution,what will be it's molarity

Chemistry

1 answer:

olga nikolaevna [1]3 years ago

6 0

<h3><u>Answer;</u></h3>

Molarity = 0.25 M

<h3><u>Explanation;</u></h3>

Molarity is given by moles/Liter.

First we find moles:

Number of moles = Mass /molar mass

= (10.7g NH4Cl)/(53.5g/mol NH4Cl)

= 0.200 moles NH4Cl

Then we convert to liters:

= (800mL)*(1L/1000mL) = 0.800L

Therefore; molarity = 0.2moles/0.8L

= 0.25M

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3 years ago

What volume of 0.152 M KMnO4 solution would completely react with 20.0 mL of 0.381 M FeSO4 solution according to the following n

ANEK [815]

<u>Answer:</u> The volume of permanganate ion (potassium permanganate) is 10.0 mL

<u>Explanation:</u>

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)

Molarity of ferrous sulfate solution = 0.381 M

Volume of solution = 20.0 mL = 0.020 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$0.381M=\frac{\text{Moles of ferrous sulfate}}{0.020L}\\\\\text{Moles of ferrous sulfate}=(0.381mol/L\times 0.020L)=0.00762mol$

For the given chemical equation:

$5Fe^{2+}+8H^++MnO_4^-\rightarrow 5Fe^{3+}+Mn^{2+}+4H_2O$

By Stoichiometry of the reaction:

5 moles of iron (II) ions (ferrous sulfate) reacts with 1 mole of permanganate ion (potassium permanganate)

So, 0.00762 moles of iron (II) ions (ferrous sulfate) will react with = $\frac{1}{5}\times 0.00762=0.00152mol$ of permanganate ion (potassium permanganate)

Now, calculating the volume of permanganate ion (potassium permanganate) by using equation 1, we get:

Molarity of permanganate ion (potassium permanganate) = 0.152 M

Moles of permanganate ion (potassium permanganate) = 0.00152 mol

Putting values in equation 1, we get:

$0.152mol/L=\frac{0.00152mol}{\text{Volume of permanganate ion (potassium permanganate)}}\\\\\text{Volume of permanganate ion (potassium permanganate)}=\frac{0.00152mol}{0.152mol/L}=0.01L=10.0mL$

Hence, the volume of permanganate ion (potassium permanganate) is 10.0 mL

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3 years ago