A 950.0 mL solution of potassium permanganate was made by dissolving 45.0 g KMnO4 in 750.0 mL of water. Assume water has a density of 1.00 g/mL. (Molar Mass for H = 1, O = 16, K = 39, Mn = 55) The mole fraction of the solute =

The correct answer is 6.67*10⁻³.

Explanation:

Based on the given question, the amount of solute (KmNO4) is 45 grams. The molecular weight of KmNO4 is 158 gram per mole. The moles of solute can be determined by using the formula,

n = mass/molecular weight

n = 45/158 = 0.28

The amount of solvent (water) given is 750 milliliters, and the density of water is 1 gm. per ml, 18 gram per mole is the molecular weight of water. So, the moles of solvent will be,

n = 750/18 = 41.7

The formula for calculating mole fraction is,

Mole fraction = mole of solute / (mass of solute + mole of solvent)

The mole fraction of solute can be determined by putting the values in the above mentioned formula,

Mole fraction of KmNO4 = 0.28 / (0.28+41.7)

= 0.28/41.98

= 6.67 * 10⁻³ or 7 * 10⁻³.