Mole fraction of liquid A (M = 20) present with liquid B(M = 25) in a solution of A (l) and B (l) is 0.6, then weight of solution per 50 g of liquid B is :
A. 55g
B. 110g
C. 220g
D. 150g
Correct Answer - B
`n_(A)=6 " mas of "A = 6xx20 =120 g`
`n_(B)=4 " mas of "B = 4xx25 =100 g`
`100 g " of "B = 220 g ` of solution
50 g of B = 110 g of solution
Correct Answer - C
`DeltaG^(@) =- "RT In K"_(eq)`
`or,- 1743 = - 8.3 xx300 xx In K_(ep)`
`K_(eq) = 2 =K_(eq) " for reaction 2A"hArr B+C " from given data"`
Calculation of molarity of the solution
Mass of NaOH =4g, Molar mass of NaoH=40 g `mol^(-1)`
Volume of solution=1 decilitre=0.1L
Molarity (M)`=("Mass of NaOH"//"Molar mass of NaOH")/("Volme of solution in...
Let the solution contain 5.0g of `Na_(2)S_(2)O_(3)` and 95.0 g of water. The number of moles of the two constitution are:
` "No. of moles of "Na_(2)S_(2)O_(3)=("Mass of" Na_(2)S_(2)O_(3))/("Molar mass") =((5.0g))/((158.0g...
Correct Answer - Solution is not ideal
The mole fraction of ethanol = 0.25
The mole fraction of water = 1-0.25=0.75
In this solution. Water is solute and alcohol is solvent....
Correct Answer - 2227.6 `Nm^(-2)`
`"Moles of urea "(n_(B))=((3g))/((60" g mol"^(-1)))=0.05 mol`
`"Moles of urea "(n_(A))=((50g))/((18" g mol"^(-1)))=2.778 mol`
`"Mole fration of water "(x_(A))=(n_(A))/(n_(A)+n_(B))=((2.778 mol))/((2.778 mol+0.05 mol))=0.9823`
`"Vapour pressure of solution"(P)=P_(A)^(@)x_(A)`...
Correct Answer - A-q ; B-p,t ; C-r,s ; D-p
(A)`P_A^0gt P_B^0` means A is more volatile than B and therefore A will be richer in vapour phase , [or `Y_A=X_A(P_A^0)/P_T`]...
Mole fraction of urea =0.4
0.4 moles of urea +0.6 moles of `H_(2)O`
`%(w//w)=(0.4xx60)/((0.4xx60)+(0.6xx18))=(24)/(24+10.8)xx100`
`=(24)/(34.8)xx100=69%`