When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi

Question

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When 16.0 g of an unknown compound (a nonelectrolyte) are dis solved in exactly 800. g of water, the solution has a freezing poi

nt of – 0.14 ° C. What is the molar mass of the compound? K f of water is 1.86 ° C m – 1 . A. 266 g B. 343 g C. 178 g D. 99.2 g E. 426 g _________________________________________________________

Chemistry

1 answer:

dexar [7] · Answer 1 · 2022-09-08T14:42:52+03:00

Answer:

A. 266g/mol

Explanation:

A colligative property of matter is freezing point depression. The formula is:

ΔT = i×Kf×m (1)

Where:

ΔT is change in temperature (0°C - -0,14°C = 0,14°C)i is Van't Hoff factor (1 for a nonelectrolyte dissolved in water), kf is freezing point molar constant of solvent (1,86°Cm⁻¹) and m is molality of the solution (moles of solute per kg of solution). The mass of the solution is 816,0g

Replacing in (1):

0,14°C = 1×1,86°Cm⁻¹× mol Solute / 0,816kg

0,0614 = mol of solute.

As molar mass is defined as grams per mole of substance and the compound weights 16,0g:

16,0g / 0,0614 mol = 261 g/mol ≈ A. 266g/mol

I hope it helps!