The mass in grams of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as produced by dissolving 1 g of urea in the same quantity of water is:

Open in App

Solution

$\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{2}}{n_{1}} = \frac{w_{2} M_{1}}{w_{1} M_{2}}$
As, $\frac{P^{0} - P s}{P^{0}}$ is the same in both cases.

${(\frac{w_{2} M_{1}}{w_{1} M_{2}})}_{g l u c o s e} = {(\frac{w_{2} M_{1}}{w_{1} M_{2}})}_{U r e a} \frac{w_{2} \times 18}{50 \times 180} = \frac{1 \times 18}{50 \times 60} w_{2} = 3 g$

Suggest Corrections

Similar questions

Q. The mass of glucose that would be dissolved in

50

g of water in order to produce the same lowering of vapour pressure as is produced by dissolving

1

g of urea in the same quantity of water is?

Q. The mass of glucose that should be dissolved in

100 g

water in order to produce same lowering of vapour pressure as produced by dissolving

1 g

urea in

50 g

of water is?

Q. The amount 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 is:

I 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 moral aqueous urea solution

Q. 18.2 g of urea is dissolved in 100g of water at

50^{o} C

. The lowering of vapour pressure produced is 5 mm Hg. Calculate the molecular mass of urea. The vapour pressure of water at

50^{o} C

is 92 mm Hg.

Join BYJU'S Learning Program

The mass in grams of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as produced by dissolving 1 g of urea in the same quantity of water is:

P0−PsP0=n2n1=w2M1w1M2 As, P0−PsP0 is the same in both cases. (w2M1w1M2)glucose=(w2M1w1M2)Ureaw2×1850×180=1×1850×60w2=3 g