Question

a solution containing 1.21 gram of camphor (molar mass is 152 gram mole inverse) in 26.68 acetone boils at 329.95 kelvin the boiling point of pure Acetone is 329. 45 Kelvin calculate molar elevation constant for acetone​

Answer 1

Answer:

Kf=1.63

If you like the answer please make as a brainliest answer

Answer 2

The molar elevation constant for acetone​ is 1.68 K/m

Explanation:

$T_b-T^o_b=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = 329.95 K

$T^o_b$ = boiling point of pure acetone = 329.45 K

$k_b$ = boiling point constant  of acetone = ?

m = molality

i = Van't Hoff factor = 1 (for non-electrolyte)

$w_2$ = mass of solute (camphor) = 1.21 g

$w_1$ = mass of solvent (acetone) = 26.68 g

$M_2$ = molar mass of solute (camphor) = 152 g/mol

Now put all the given values in the above formula, we get:

$(329.95- 329.45)K=1\times k_b\times \frac{(1.21g)\times 1000}{152\times (26.68g)}$

$0.50K=1\times k_b\times 0.298m$

$k_b=1.68K/m$

Learn more about elevation in boiling point

https://brainly.com/question/8210955

https://brainly.in/question/10647471