Question

What is the concentration of sugar in mol/L if its 20 g are dissolved in enough water to make a final volume up to 2L?​

Answer 1

Answer :

• Concentration of sugar = 0.0293 M.

Step-by-step explanation :

Given that,

• Weight of solute, W = 20 g.
• Volume of solution, V = 2 L.

$\hrulefill$

Molar mass of sugar $\rm{(C_{12}H_{22}O_{11})}$ =

$\implies\rm{12\times{}12+22\times{}1+11\times{}16}$

$\implies\rm{342\:g\:mol^{-1}}$

Now,

No. of moles of sugar = $\rm{\dfrac{W}{M}=\dfrac{20}{342}}$

$\implies\rm{0.0585\:mole}$

∴ Molar concentration (molarity) = $\rm{\dfrac{n}{V}}$

Putting the values we found, we get,

$\implies\rm{\dfrac{0.0585}{2}=}\:\bold{0.0293\:M}$

Answer 2

Explanation:

Concentration of sugar = 0.0293 M.

Step-by-step explanation :

Given that,

Weight of solute, W = 20 g.

Volume of solution, V = 2 L.

Molar mass of sugar \rm{(C_{12}H_{22}O_{11})}(C

12

H

22

O

11

) =

\implies\rm{12\times{}12+22\times{}1+11\times{}16}⟹12×12+22×1+11×16

\implies\rm{342\:g\:mol^{-1}}⟹342gmol

−1

Now,

No. of moles of sugar = \rm{\dfrac{W}{M}=\dfrac{20}{342}}

M

W

=

342

20

\implies\rm{0.0585\:mole}⟹0.0585mole

∴ Molar concentration (molarity) = \rm{\dfrac{n}{V}}

V

n

Putting the values we found, we get,

\implies\rm{\dfrac{0.0585}{2}=}\:\bold{0.0293\:M}⟹

2

0.0585

=0.0293M