Question

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Answer 1

Answer:

$\boxed{\underline{\huge{\texttt{x = 36.946 g}}}}$

Explanation:

Given,

Molality (m) = $0.25 \text{ mol kg}^{-1}$

Mass of the solution = 2.5 kg = 2500 g

To find,

Mass of urea required = ?

 

[Molality = given mass of solute/molar mass × 1000/mass of the solvent (in g)]

Molar mass $=(14 + 2 + 12 + 16 + 14 + 2) = 60\text{ g mol}^{-1 }$

Let x be the mass of urea required.

Mass of the solvent = (2500-x)

Substituting the values,

$0.25 = \frac{x}{60} \times \frac{1000}{(2500-x)}\\\\(2500-x) = \frac{x}{60} \times \frac{1000}{0.25}\\\\(2500-x) = \frac{200x}{3}\\\\2500=\frac{203x}{3}\\\\x=\frac{7500}{203}\\\\x=36.946 g$

Answer 2

Answer:

Answer:

\boxed{\underline{\huge{\texttt{x = 36.946 g}}}}

x = 36.946 g

Explanation:

Given,

Molality (m) = 0.25 \text{ mol kg}^{-1}0.25 mol kg

−1

Mass of the solution = 2.5 kg = 2500 g

To find,

Mass of urea required = ?

[Molality = given mass of solute/molar mass × 1000/mass of the solvent (in g)]

Molar mass =(14 + 2 + 12 + 16 + 14 + 2) = 60\text{ g mol}^{-1 }=(14+2+12+16+14+2)=60 g mol

−1

Let x be the mass of urea required.

Mass of the solvent = (2500-x)

Substituting the values,

\begin{gathered}0.25 = \frac{x}{60} \times \frac{1000}{(2500-x)}\\\\(2500-x) = \frac{x}{60} \times \frac{1000}{0.25}\\\\(2500-x) = \frac{200x}{3}\\\\2500=\frac{203x}{3}\\\\x=\frac{7500}{203}\\\\x=36.946 g\end{gathered}

0.25=

60

x

×

(2500−x)

1000

(2500−x)=

60

x

×

0.25

1000

(2500−x)=

3

200x

2500=

3

203x

x=

203

7500

x=36.946g