Question

A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.

NCERT Class XII
Chemistry - Main Course Book I

Chapter 2. Solutions

Answer 1

mass of solution of 300 gm of 25%= 300x25/100= 75 gm                                      mass of solution of 400 gm of 40% = 400x40/100= 160 gm                                                                total of above  =(160+75)= 235 gm                                       in 700 gm solut. the mass is= 235 gm
in  1 gm    solut. the mass is=235/700 gm
in 100 gm solut. the mass is=235x100/700 gm
                                         =33.57 gm or 33.75%              

Answer 2

Let A be the first solution and B is second solution.

★Given that:

• Mass of Solute in solution A, Ma = [25 × 300]/100 = 75g
• Mass of Solute in solution B, Mb = [40 × 400]/100. = 160g

†To find:

• Mass Percentage of Solute

†Formula:

Mass Percentage of Solute

$= \frac{mass \: of \: solute}{mass \: of \: solution} \times 100$

★SolutiOn:

⇒ Total mass of Solute in mixture, M = Ma + Mb

= [75 + 160]

= 235g

⇒ Total mass of solution = 300 + 400

= 700g

⇒ Mass Percentage of Solute

= 235/700×100

= 33.5%

⇒ Percentage of Water in final solution = 100-33.5

= 66.5%

Therefore mass percentage of solute is 33.5% and that of water is 66.5%.

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