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2 years ago

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A pipet is used to transfer 5.00 mL of a 1.25 M stock solution in flask "S" to a 25.00 mL volumetric flask "B," which is then di

luted with DI H2O to the calibration mark. The solution is thoroughly mixed. Next, 2.00 mL of the solution in volumetric flask "A" is transferred by pipet to 50.00 mL volumetric flask "B" and then diluted with DI H2O to the calibration mark. Calculate the molarity of the solution in volumetric flask "B". How do I solve this?

1 answer:

Leto [7]2 years ago

5 0

Answer: the molarity of the solution in volumetric flask "B' is 0.0100 M

Explanation:

Given that;

the Molarity of stock solution M₁ = 1.25M

The molarity os solution in volumetric flask A (M₂) = M₂

Volume of stock solution pipet out (V₁) = 5.00mL

Volume of solution in volumetric flask A V₂ = 25.00mL

using the dilution formula

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

WE SUBSTITUTE

M₂ = ( 1.25 × 5.00 ) / 25.00 mL

M₂ = 0.25 M

Now volume of solution pipet out from volumetric flask A V₂ = 2.00 mL

Molarity of solution in volumetric flask B (M₃) = M₃

Volume of solution in volumetric flask B V₃ = 50.00m L

Using dilution formula again

M₂V₂ = M₃V₃

M₃ = M₂V₂ / V₃

WE SUBSTITUTE

M₃ = ( 0.25 × 2.0) / 50.0

M₃ = 0.0100 M

Therefore the molarity of the solution in volumetric flask "B' is 0.0100 M

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Data Given:

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many grams of NH₃ = ?

Solution:

First we look at the balanced synthesis reaction

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1 mol 3 mol

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Convert moles to mass

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molar mass of H₂ = 2(1) + 2 g/mol

Now

N₂ + 3 H₂ ------—> 2 NH₃

1 mol (28 g/mol) 3 mol(2g/mol)

28 g 6 g

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Apply Unity Formula

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Do cross multiply

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To find mass of NH₃ we will do following calculation

Look at the reaction

As we Know

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Apply Unity Formula

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Do cross multiply

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Formula will be used

mass in grams = no. of moles x molar mass . . . . . . (2)

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