The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations. rank the solutions according to their respective osmotic pressures in decreasing order assuming the complete dissociation of ionic compounds. rank from highest to lowest osmotic pressure. to rank items as equivalent, overlap them. 1m c6h12o6, 1m mgcl2, 2m ch3oh, 1m nacl

C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂

Step-by-step explanation:

The formula for osmotic pressure (Π) is

Π = icRT

If T is constant,

Π = kic

C₆H₁₂O₆:

Π = k × 1 × 1 = k

MgCl₂:

Π = k × 3 × 1 = 3k

CH₃OH:

Π = k × 1 × 2 = 2k

NaCl:

Π = k  × 2 × 1 = 2k

The order of osmotic pressures is  

C₆H₁₂O₆ < NaCl ≈ CH₃OH < MgCl₂

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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