Question

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

Answer 1

0.50M of solution

Step-by-step 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

0.5 M

Step-by-step explanation:

First we have to start with the molarity equation:

`M=(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(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(1~mol~ Na_2SO_4)/(142~g~ of~Na_2SO_4)`

`0.05~mol~ Na_2SO_4`

Finally, we can calculate the molarity:

`M=(0.05~mol~ Na_2SO_4 )/(0.1~ L)`

`M=0.5`

I hope it helps!