Home
Class 12
CHEMISTRY
An aqueous solution of glucose was prepa...

An aqueous solution of glucose was prepared by dissolving 18g of glucose in 90g of water. The relative lowering in vapour pressure is

A

0.02

B

1.1

C

20

D

196

Text Solution

Verified by Experts

The correct Answer is:
A
`(DeltaP)/(p^(@)) = (18 xx 18)/(180 xx 90) = 0.02`
Promotional Banner

Topper's Solved these Questions

  • SOLUTIONS

    DINESH PUBLICATION|Exercise REVISION QUESTION|186 Videos

  • SOLUTIONS

    DINESH PUBLICATION|Exercise OBJECTIVE TYPE MCQs|47 Videos

  • SOLID STATE

    DINESH PUBLICATION|Exercise Brain storming|10 Videos

  • SOME BASIC CONCEPTS OF CHEMISTRY

    DINESH PUBLICATION|Exercise ULTIMATE PREPARATORY PACKAGE|20 Videos

Similar Questions

Explore conceptually related problems

3g urea is dissolved in 45g of water. The relative lowering of vapour pressure is :

6g of urea is dissolved in 90g of water. The relative lowering of vapour pressure is equal to

(a) A solution is prepared by dissolving 10g of non-volatile solute in 180g of H_(2)O . If the relative lowering of vapour pressure is 0.005 , find the molar mass of the solute. (b) How many grams of sucrose (C_(12)H_(22)O_(11)) must be dissolved in 90g of water to produce a solution over which the relative humidity is 80% ? Assume the solution is ideal.

S34.2 g of cane sugar is dissolved in 180 g of water. The relative lowering of vapour pressure will be

An aqueous solution of glucose is made by dissolving 10 g of glucose in 90 g water at 303 K. If the V.P. of pure water at 303 K be 32.8 mm Hg, what would be V.P. fo the solution ?

At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea =60 g "mol"^(-1) )