Question

1 M of glucose (CH20) solution (density = 1.18 g/ml) is equivalent to which of the following solution
(A) % w/w = 18% (solution)
(B) 180 g solute
per
litre solution
(C)% w/v = 18% (solution)
(D) 1 molal solution​

Answer 1

Given : molarity of glucose in solution is 1M and density of solution is 1.18 g/ml.

To check : (A) % w/W = 18%

(B) 180g solute per litre solution.

(C) % w/V = 18 %

(D) 1 molal solution.

solution : molarity of glucose in solution is 1M. it means one mole of glucose (solute) presents in one litre of solution.

mass of one mole of glucose = 180 g [ as molecular weight of Glucose is 180g/mol]

so, 180g of solute per litre of solution, is correct statement.

mass of solution = volume × density

= 1000ml × (1.18 g/m)

= 1180 g

mass of solvent = 1180g - 180g = 1000g

percentage by weight , % w/W = 180/1180 × 100

= 15.25 %

hence, % w/W = 18 % , is incorrect statement.

now percentage weight by volume, % w/V = mass of solute/volume of solution × 100

= 180g/1000 ml × 100

= 18 %

so, % w/V = 18% , is a correct statement.

molality = no of moles of solute/mass of solvent

= 1/1000 × 100

= 0.1 molal

so, 1 molal solution, is incorrect statement.

Therefore options (B) and (C) are correct choices.

Answer 2

opt d is also crt

solvent=1000g=1kg

molality=moles / weight of solvent in kg

=1/1=1 molal