Question

500 gm of urea solution of mole fraction 0.2 is diluted to 1500gm .calculate the mole fraction of solute in diluted solution

Answer 1

Answer: Mole fraction of Urea in diluted solution 0.05 .

Solution:

Let the amount of urea present in total 500 g solution be x grams

The amount of water present in 500 g of solution be (500-x) grams

$X_{urea}=o.2=\frac{\text{moles of urea}}{\text{number of moles of urea+ number of mole of water}}=\frac{\frac{x}{60}}{\frac{x}{60}+\frac{500-x}{18}}}$

Solving for x we get :

$x= 227.27 g$ that is mass of urea preseni in 500 grams od solution

The amount of water present in 500 g of solution = (500-x) = 272.73 grams

After dilution solution weighed 1500 grams means that 1000 g of water was added to 500 g of solution :

New mass of the water(solvent) in 1500 g solution : 1000 g+ 272.73 g= 1272.73 g

New mass of the urea(solute) in 1500 g solution will remain same as in 500 g solution: 227.27 g

Mole fraction of Urea in diluted solution:

$X'_{urea}=\frac{\frac{227.27 g}{60 g/mol}}{\frac{227.27g}{60 g/mol}+\frac{1272.73 g}{18g/mol}}=0.05$

Answer 2

Answer:

mole fraction of urea in diluted solution 0.05