How many grams of KCI are needed to prepare 2.0 liters of 2.0 M solution?

Molecular weight of KCl is , 1x39 + 1x 35.5 = 74.5 gram/mole, since 1 molar solution of any compund is equal to molecular weight of the compound in grams present in 1 litre of water , so 1 M solution of KCl is 74.5 grams in 1 liter volume and therefore 2M solution would correspond to 2x 74.5 grams per litre volume , so 2 M KCl solutions of 2 litres would correspond to 2litres x 2 x 74.5 grams /litre = 298 grams of KCl

Equation:

Concentration = moles/ volume (in litres)

Molar mass of KCl = 35.5 + 74.5 = 74.5g

If you rearrange formula, you should get:

Concentration x volume = moles

So, 2 x 2 = 4

4 moles needed = 4 * 74.5g = 298g

For 1litre solution or 1000ml

Molarity = (mass/molar mass) x (1/1000)

For 2 liter or 2000ml solution

Molarity = (mass/molar mass) x (1000/2000)

Molarity = (mass/molar mass) x 0.5

Now rearrange above relation to calculate mass

Mass =( molarity x molar mass) / 0.5

Mass = (2.0 x 74.55)/ 0.5

Mass = 298.2 gram

298.2 gram of KCl is required to prepare 2.0M solution

Find the number of moles of KCl needed:

M = n/V

2.0M = n/2.0L

n = 4.0 mol

Find the number of grams of KCl in 4 moles of KCl:

n = m/M

4.0mol = m / 74.5483g/mol

m = 298.1932g

m = 3.0 x 10^2g (2 Significant figures)

Therefore 3.0x10^2 grams of KCl is needed.

Hope this helps!

This equation:

MV = mass / molar mass

should be used.

(2.0 mol/L) (2.0 L) = x / 74.551 g/mol

x = 298.204 g (round off more as you see fit)