ExamHabit
Practise free

JEE Main 2019 · 12 Jan · Shift II

JEE Main 2019 Chemistry question on Solutions

The amount of sugar (C12H22O11\text{C}_{12}\text{H}_{22}\text{O}_{11}) required to prepare 2 L of its 0.1 M aqueous solution is
  1. A.
    17.1 g
  2. B.
    68.4 g
  3. C.
    136.8 g
  4. D.
    34.2 g
Show answer and solution

Correct option: B

Solution

Moles needed = Molarity × Volume = 0.1 mol/L × 2 L = 0.2 mol. Molar mass of C₁₂H₂₂O₁₁ = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 g/mol. Mass = 0.2 × 342 = 68.4 g.

[JEE Main 2019 · 12 Jan · Shift II]

Practise Solutions the way it is asked

A timed set of real JEE Main Solutions questions, graded as the exam grades. No sign-up.

Start a Solutions test

More: all JEE Main Solutions questions · JEE Main 2019 · Chemistry chapter list