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3 years ago

Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water

Chemistry

1 answer:

babymother [125]3 years ago

6 0

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/ 53.491 g/mol

Number of moles = 0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality = 0.12 mol / 0.016 kg

Molality = 7.5 m

or (m=mol/kg)

Molality = 7.5 mol/kg

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Answer:

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Explanation:

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21.1 g / 36.45 g/mol = 0.579 moles of HCl

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4 years ago

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3 years ago

The pOH of a solution of KOH is 11.30. What is the [H*] for this solution?

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Explanation:

Given: pOH = 11.30

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pH = 14 - 11.30

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$pH = - log [H^{+}]$

Substitute the values into above formula as follows.

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3 years ago