Question

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

Answer 1

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) = 0.05 moles of Na₂SO₄.

In 100mL = 0.100L:

0.05 moles Na₂SO₄ / 0.100L = 0.50M of solution

Answer 2

Answer:

0.5 M

Explanation:

First we have to start with the molarity equation:

$M=\frac{mol}{L}$

We need to know the amount of moles and the litters.

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 formula of sodium sulfate $Na_2SO_4$, with this formula we can calculate the molar mass 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 calculate the moles:

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

$0.05~mol~ Na_2SO_4$

Finally, we can calculate the molarity:

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

$M=0.5$

I hope it helps!