Question

the amount of glucose to be dissolved in 500g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution is​

Answer 1

Explanation:

For urea :m=0.2 and m=moles of urea(n)*1000/mass of solventSo, n*1000/mass of solvent=0.2By this, relative lowering in vapour pressure =n*molar mass of solvent/mass of solvent=0.2*18/1000=3.6/1000 .....(1)For glucose : relative lowering in vapour pressure =n*molar mass of solvent(water)/mass of water=mass of glucose (W)*18/180*500 =W/5000 ........(2)From 1and2,3.6/1000=W/5000So W=18g

hope it helps you pls mark me as brainlist answer

Answer 2

180 grams of glucose to be dissolved in 500 g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution

Explanation:

Molality of the urea solution = 0.2 mol/kg

This means that 0.2 moles of urea is present in 1000 g of the water.

Moles of water = $\frac{1000 g}{18 g/mol}=55.55 mol$

Moles of urea = 0.2 mol

Mole fraction of urea = $\chi_1=\frac{0.2 mol}{0.2 mol+55.55 mol}=0.003587$

Relative lowering in vapor pressure :

$\frac{p^o-p_s}{p^o}=i\times \chi_{solute}$

where,

= relative lowering in vapor pressure

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

$\chi_{solute}$ = mole fraction of solute

$p^o$ = vapor pressure of pure solvent=

$p_s$ = vapor pressure of solution =

Moles of glucose to be added in 500 g of water = n

Moles of water in glucose solution = $=\frac{500 g}{18 g/mol}=27.78 mol$

Mole fraction of glucose = $\chi_2=\frac{n}{n+27.78 mol}$

$\chi_1=\chi_2$

$0.003587 =\frac{n}{n+27.78 mol}$

n = 0.100 mol

Mass of 0.100 mole of glucose = :

0.100 mol × 180 g/mol = 18 g

180 grams of glucose to be dissolved in 500g of water so as to produce the same lowering in vapour pressure as that of 0.2 molal aqueous urea solution.

Learn more about : Relative lowering in vapor pressure:

https://brainly.com/question/14121655

https://brainly.in/question/14867528

#learnwithbrainly