Question

18 g of glucose is present in 100 mL of its
solution. The concentration of solution in
mol L-1 will be
(2) 0.1 mol L-1
(1) 2 mol L-1
(4) 0.2 mol L-1
(3) 1 mol L-1​

Answer 1

Answer:

1 mol/ltr or 1M

Explanation:

C6H12O6 molar mass= 6(12)+12+(16)6

=180g

given mass of solute=18g

vol of solution=100ml

Molarity=given mass/ molar mas×1000/vol of solution in ml

= 18/180×1000/100

= 1M

Hence option 3 is correct 1mol L^-1

Answer 2

Answer:

$\bf{1\:mol\:L^{-1}}$

Explanation:

Mass of glucose (Solute) = 18 g.

No. of moles = Mass/Molar mass = 18/180 = 0.1 mole

Total volume of the solution = 100 mL = 0.1 L

Molar concentration = Number of moles of solute/Volume of solution

$\longrightarrow\tt{Molar\:concentration=\dfrac{0.1\:mol}{0.1\:L}}$

$\longrightarrow\bf{1\:mol\:L^{-1}}$

Some important formulas :

• Number of moles = Mass/Molar mass
• Molar concentration = Number of moles of solute/Volume of solution
• Mass % = Mass/Molar mas * 100 %
• n = Molecular mass/Empirical Formula mass
• Mole Fraction : X_A = n_A/(n_A+n_B) or X_B = n_B/(n_A+n_B)
• Molarity = M = n_solute/1L volume of soln
• Molality = m = n_solute/1 kg of soln