Question

6.02*10^21 molecules of ammonia are present in 250 mL of its solution. Molarity of the solution is​

Answer 1

Answer:

212.98

Explanation:

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Answer 2

The molarity of the solution is 0.4M

GIVEN

Number of molecules of ammonia = 6.02 × 10²¹

Volume of ammonia solution = 250 ml

TO FIND

We can simply solve the above problem as follows;

We know that,

Molarity, M = Moles of solute/Volume of solution in Liters.

It is given that,

Number of Molecules of Ammonia = 6.02 × 10²¹

We know that 1 mole of substance = 6.02 × 10²³ molecules of substance.

Moles of Ammonia =

$= \frac{1}{6.02 \times {10}^{23} } \times 6.02 \times {10}^{21}$

= 10²¹ × 10⁻²³

= 10⁻²

= 0.01 moles.

Volume of solution = 250 ml = 0.025 litres.

Molarity = 0.01/0.025 = 0.4 M

Hence, The molarity of the solution is 0.4M

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