Question

What is the molarity of 225 mL solution containing 1.89g of K₂SO₃•8H₂O? * 2 points 3.77 x 10⁻³ M 6.44 x 10⁻¹ M 3.87 x 10⁻²M 2.78 x 10⁻² M 1.18 x 10⁻³ M

Answer 1

Answer:

2.78 x 10⁻² M

Explanation:

Step 1:

Data obtained from the question.

Volume = 225mL / 1000 = 0.225L

Mass of K₂SO₃•8H₂O = 1.89g

Molarity =..?

Step 2:

Determination of the number of mole of K₂SO₃•8H₂O

This is illustrated below:

Mass of K₂SO₃•8H₂O = 1.89g

Molar mass of K₂SO₃•8H₂O = (39x2) + 32 + (16x3) + 8[(2x1) + 16] = 78 + 32 + 48 + 8[18] = 302g/mol

Number of mole = Mass/Molar Mass

Number of mole of K₂SO₃•8H₂O = 1.89/302 = 6.26x10⁻² mole

Step 3:

Determination of the molarity.

Molarity = mole /Volume

Molarity = 6.26x10⁻² /0.225

Molarity = 2.78 x 10⁻² M

Therefore, the molarity of the solution is 2.78 x 10⁻² M