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3 years ago

How many grams of nacl are required to make 250.0 ml of a 3.000 m solution?

1 answer:

8 0

Molarity (M) = moles of solute (mol) / Volume of the solution (L)

Molarity of the solution = 3.000 M
Volume of the solution = 250.0 mL = 0.25 L
moles in 250.0 mL = molarity x volume of the solution
= 3.000 M x 0.25 L
= 0.75 mol

Hence, 0.75 mol of NaCl is needed to prepare 250.0 mL of 3.000 M NaCl solution.

Moles (mol) = mass (g) / molar mass (g/mol)

Moles of NaCl in 250.0 mL = 0.75 mol
Molar mass of NaCl = 58.44 g/mol
Mass of NaCl in 250.0 mL = Moles x Molar mass
= 0.75 mol x 58.44 g/mol
= 43.83 g

Hence, 43.83 g of NaCl is needed to prepare 250.0 mL of 3.000 M solution.

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The law of conservation of mass was obeyed since some of the mass evaporated while the cookies were baking.

Explanation:

Maria mixes water, flour, sugar, baking powder, butter and chocolate chips and then measures the mass of her ingredients.

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All these information are not of any great help.

Law of conservation of mass states that:

For any system closed to all transfers of matter and energy, the mass of the system can not change with respect to time, as system's mass cannot change, any mass can neither be added nor be removed.

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a sealed container filled with argon gas at 35 c has a pressure of 832 torr. if the volume of the container is decreased by a fa

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Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m i

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First, we need to get the value of Ka:

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by using the ICE table :

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change -X +X +X

Equ (0.013-X) X X

when Ka = [NH3][H+] / [NH4+]

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