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Solutions

Question

6.02×10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of urea solution is

0.001 M

0.1 M

0.02 M

0.01 M

Solution

0.01 M

Avogadros number N_A = 6.02 x 10²³ = 1 mole
∴ 6.02 x 10²⁰ molecules
= 0.001 mole in 100 mL (0.1 L solution)
∴ Molar concentration = mol/volume in L
=0.001/0.1
= 0.01 M

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Solutions

6.02×10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of urea solution is

0.001 M

0.1 M

0.02 M

0.01 M

Some More Questions From Solutions Chapter

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 37⁰C.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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Solutions

6.02×1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is 0.001 M 0.1 M 0.02 M 0.01 M

Some More Questions From Solutions Chapter

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 370C.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

6.02×10²⁰ molecules of urea are present in 100 ml of its solution. The concentration of urea solution is

0.001 M

0.1 M

0.02 M

0.01 M

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL at 37⁰C.