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4 years ago

13

Which of the following should be classified as a solution?

2 answers:

Westkost [7]4 years ago

7 0

Answer:

a soft drink with no fizz

Valentin [98]4 years ago

4 0

A soft drink with no fizz would be the answer because in a solution there's a solute, which makes it a little bit of a solution; it is usually dissolved. (Which is the sugar/sodium in a soft drink. Then, the solvent makes up most of the solution, which is water in this case.

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A certain substance X condenses at a temperature of 123.3 degree C . But if a 650. g sample of X is prepared with 24.6 g of urea

Pavel [41]

Answer:

<u>The molal boiling point elevation constant is 1.59 ≈ 1.6</u> $Kkgmol^{-1}$

Explanation:

To solve this question , we will make use of the equation ,

Δ $T_{b} = i*K_{b} *m$

<em>where , </em>

<em>Δ $T_{b}$ is the change in boiling point of the substance $X$ ( ° $C$ or $K$ )</em>

<em> $i$ is the Vant Hoff Factor which = 1 in this case ( no unit )</em>

<em> $K_{b}$ is the mola boiling point elevation constant of X ( $Kkgmol^{-1}$ )</em>

<em> $m$ is the molality of the solution which has $(NH_{2})_{2} CO$ as the solute and $X$ as the solution ( $molkg^{-1}$ )</em>

Δ $T_{b}$ = $124.3 -123.3 = 1$ ° $C$ or $K$ ;
$i$ =1;
$m$ = $\frac{moles of solute}{weight of solvent(kg)}$ $molkg^{-1}$

∴ $m = \frac{\frac{24.6}{60} }{\frac{650}{1000} }$

<em>as the weight of $(NH_{2})_{2} CO$ is $60g$ and thus number of moles = $\frac{24.6}{60}$ </em>

<em>and the weight of solvent in $kg$ is $\frac{650}{1000}$ </em>

4. $K_{b}$ ⇒ ?

∴

$1=1*K_{b} *\frac{\frac{24.6}{60} }{\frac{650}{1000} }$

⇒ $K_{b}$ = $1.59$ ≈ 1.6 $Kkgmol^{-1}$

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