Test: Basic Concepts of Solutions(17 Sep) - JEE MCQ

At dynamic equilibrium, number of solute particles going into the solution will be equal to solute particles separating out. Hence, the concentration of solute in the solution remains constant.

p = K_Hx. Higher the value of K_H at a given pressure, lower is the solubility of the gas in the liquid.

No. of moles of H₂S = 0.195
No. of moles of H₂O = 1000/18 = 55.55mol
Mole fraction of H₂S =
Pressure at STP = 0.987 bar
According to Henry’s law, p = K_H^x
or K_H = pH₂S/xH₂S = 0.98/70.0035
= 282 bar

p = K_H × x

Mole fraction = Moles of gas/Total moles
Moles of H₂O = 1000/18 = 55.55 (∵ 1L = 1000 g)
Mole fraction = (55.55 >>> x)

No. of moles of glucose = 10/180 = 0.0555 mol
No. of moles of water = 90/18 = 5 mol
Number of moles of solution = 5.0555 mol
Mole fraction of glucose = No. of moles of glucose/No. of moles of solution = 0.0555/5.0555 = 0.01

ppm = 

= 10.4 ppm

No. of moles of NaOH = 10/40 = 0.25 mol

No. of moles of NaOH

Mass of NaOH = 40 × 0.125 = 5g

Mass of solute taken = 120g
Molecular mass of solute = 60u
Mass of solvent = 1000g
Density of solution = 1.15g/mL
Total mass of solution = 1000 + 120 = 1120g

Molality = no of moles of solute/mass of solvent in kg
Molar mass of water = 18 g
Mass of water = 648 g
No of moles of water = 648/18 = 36 mol