Answer:
AS both the solution are isotonic they should have same concentrationin mole/litre
for sucrose solution concentration =4g/100cm3
= 40g/Litre
molar mass of sucrose C 12H22O11 = 342
therefore we get =40/342 moles/litre
For unknown substance Let N be the molecular mass then concentration = 3g/100cm3
= 30g/Litre
= 30/N moles/Litre
comparing the both equation
30/N = 40/342
N=(30 x 342)/40
N= 256.5
Molecular mass of oragnic compound
Answer:
= xurea
Let the mass of solution be 100g
therefore mass of urea = 10 g
molecular mass of urea = 60 g
xurea = 10/60 =1/6
molecular mass of water = 18g
xwater = 90/18 =5
The relative lowering of vapour pressure is given by the following expression.
Where is the vapour pressure of pure solvent, Psolution is the vapour pressure of solution containing dissolved solute, n1 is the number of moles of solvent and n2 is the number of moles of solute.
For dilute solution n2<<n1, therefore the above expression reduces to
Where w1 and w2 are the masses and M1 and M2 are the molar masses of solvent and solute respectively.
We are given that
W1 =30 g
W2 =30g
psolution =2.8 kpa
p0solvent =? And M2=?
Answer:
Calculate the vapour- pressure lowering of water when5.67g of glucose, C6H12O6 is dissolved in 25.2 g water at 250C. The vapour pressure of water at 250C is 23.8 mmHg. What is the vapour pressure of the solution?
WB =5.67g
WA =25.2g
MB=180g/mol
MA=18g/mol
NB= 5.67/180 =0.0315 mol glucose
NA=25.2/18 =1.40 mol water