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BaLLatris [955]
3 years ago
15

The molarity of a solution containing 7.1 g of sodium sulfate in 100 mL of an aqueous solution is

Chemistry
2 answers:
zhenek [66]3 years ago
8 0

Answer:

0.50M of solution

Explanation:

Molarity is defined as the ratio between moles of solute (In this case, sodium sulfate, Na₂SO₄) per liter of solution.

Moles of 7.1g of Na₂SO₄ (Molar mass: 142g/mol) are:

7.1g Na₂SO₄ × (1mol / 142g) =<em> 0.05 moles of Na₂SO₄</em>.

In 100mL = 0.100L:

0.05 moles Na₂SO₄ / 0.100L = <em>0.50M of solution</em>

Sergio [31]3 years ago
4 0

Answer:

0.5 M

Explanation:

First we have to start with the <u>molarity equation</u>:

M=\frac{mol}{L}

We need to know the<u> amount of moles and the litters</u>.

If we have 100 mL we can convert this value to “L”, so:

100~mL\frac{1~L}{1000~mL}

0.1~ L

Now we can continue with the moles, for this we have to know the <u>formula of sodium sulfate</u> Na_2SO_4, with this formula we can <u>calculate the molar mass</u> if we know the atomic mass of each atom on the formula (Na: 23 g/mol, S: 32 g/mol, O: 16 g/mol). We have to multiply each atomic mass by the amount of atoms in the formula, so:

molar~ mass~=~ (23*2)+(32*1)+(16*4)= ~ 142~ g/mol

In other words:

1~mol~ Na_2SO_4=~142~g~ of~Na_2SO_4

Now we can <u>calculate the moles</u>:

7.1~g~ of~Na_2SO_4\frac{1~mol~ Na_2SO_4}{142~g~ of~Na_2SO_4}

0.05~mol~ Na_2SO_4

Finally, we can <u>calculate the molarity:</u>

M=\frac{0.05~mol~ Na_2SO_4 }{0.1~ L}

M=0.5

I hope it helps!

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