The amount of sugar ( C _ 12 H _ 22 O _ 11 ) required to prepare 2L of its 0 .1 M aqueous solution is:

Correct answer: C. 68 .4 g Molarity = (Moles)/(Volume(inL)) ⇒ 0.1 = (moles)/(2) Moles of sucrose required = 2 × 0.1 = 0.2 moles Weight of sucrose = 0 .2 × 342g = 68 .4 g

JEE Main 201910 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 10 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2019 · 10 January · Evening Shift · Chemistry · Solutions

The amount of sugar ( C _ 12 H _ 22 O _ 11 ) required to prepare 2L...

The amount of sugar

\left({C}_{12}{H}_{22}{O}_{11}\right)

required to prepare

2L

of its

0 .1 M

aqueous solution is:

Options

A $17 .1 g$
B $136 .8 g$
C $68 .4 g$
D $34 .2 g$

Correct answer

C. $68 .4 g$

Step-by-step solution

Molarity = \dfrac{Moles}{Volume\left(inL\right)}

<br /><br />

⇒ 0.1 = \dfrac{moles}{2}

<br /><br />Moles of sucrose required

= 2 \times 0.1 = 0.2

moles<br /><br />Weight of sucrose

= 0 .2 \times 342g$$= 68 .4 g

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