How many grams of naoh are needed to prepare 2.0 l of a 1.5 m naoh solution?

120 grams

Explanation:

Data:

Volume = V= 2 L

Molarity = 1.5 M

Molecular masses:

Na = 23, O= 16, H=1

Total molecular mass =M= 23+16+1 = 40

mass = m = ?

To find the mass we need to find moles from molarity formula:

Molarity = n/V

n= Molarity x V

= 1.5 x 2

n= 3 moles

Now we can find mass:

number of moles = given mass / molecular mass

n = m/M

m= n x M

= 3 x 40

m = 120 grams

120 grams of NaOH

Explanation:

This can be solved by using Unit Conversion.

Let n =  # grams of NaOH needed to prepare 2.0 L of a 1.5 M NaOH solution.

Known variables

V = 2.0 L

[NaOH] = 1.5 mol/L

Molecular Weight of NaOH = 39.997 g/mol

n =  2.0 L x  (1.5 mol/L) x (39.997 g NaOH/ mol) =  119.991 grams

rounding to two significant figures

n = 120 grams of NaOH

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