Question

please jldi ans kro ........................​

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:

Compounds similar to glucose or alcohol do contain hydrogen element but they do not show signs of acidic nature. The fact that the hydrogen in them will not separate as like the hydrogen in the acids. They will not separate to become hydrogen ions, on dissolving in the water.