Boiling point elevation

Chemistry Level 2

When 3.15 3.15 g of unknown hydrocarbon is dissolved in 75.0 75.0 g of benzene, the boiling point of the solution increases by 0.597 0.597 °C. What is the molar mass of the unknown substance?

The K b K_{b} of benzene is 2.53 2.53 K kg mol − 1 . ^{-1}.

86 g mol − 1 86\text{ g mol}^{-1} 121 g mol − 1 121\text{ g mol}^{-1} 178 g mol − 1 178\text{ g mol}^{-1} 242 g mol − 1 242\text{ g mol}^{-1}

Krishna Singh by Krishna Singh , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

James Wilson
Nov 6, 2017

I didn't know a single chemistry formula to help me with this, but I was able to reason this out on my own simply by thinking about it. Here's the calculation I did: ( 3.15 ) ( 2530 ) ( 75 ) ( 0.597 ) \frac{(3.15)(2530)}{(75)(0.597)} . However, it seemed pretty strange to consider K / m o l K/mol as a unit in the reasoning that I used.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Alakh Aggarwal
May 14, 2014

apply delta(T [boliling] ) = Kb m ......... and m = n[solute]* 1000/g[solvent(in grams)] ...... n[solute] = g[solute]/M[solute]to get M[solute] = 178 g/mol

0 Helpful 0 Interesting 0 Brilliant 0 Confused

If you allow me:

Δ T B = i ⋅ K b ⋅ m \Delta{}_{T_B}=i\cdot{}K_b\cdot{}m

( I won't write the units. )

0.597 = 1 ⋅ 2.53 ⋅ x 0.075 0.597=1\cdot{}2.53\cdot{}\frac{x}{0.075}

x ≅ 0.017697628458 x\cong{}0.017697628458

3.15 x = M M H C ≅ 177.9899 \frac{3.15}{x}={MM}_{HC}\cong{}177.9899

Bernardo Sulzbach - 6 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...