Question

what is the molarity of NAOH solution if 250ml of it contain 1 mg of NAOH

Answer 1

molarity= no of moles/ volume in litres
mass given= 1 mg= 10^-3 grams
no of moles= 10^-3/molar mass
molar mass= 40
n= 10^-3/40
now volume= 250 ml =250/1000 L
so M= 10^-3/40 divided by 250/1000 L
10^-3/40 x 1000/250= 10^-3/40 x 4= 10^-4 M

Answer 2

The Main Answer is: The molarity of the given NaOH solution is $10^{-4}$ M.

Given: Volume of NaOH solution = 250 ml

           Mass of NaOH = 1 mg = $10^{-3}$ gm

To Find: Molarity of the NaOH solution

Solution:

We know,

Atomic mass of Na = 23 gm

Atomic mass of O = 16 gm

Atomic mass of H = 1 gm

Therefore, the molecular mass of NaOH = 23 + 16 + 1 = 40 gm

We know,

Molarity of a solution = $\frac{No. of moles of solute}{Total volume of solution in ml}$ × 1000

Here, NaOH is the solute.

Now, no. of moles of NaOH = $\frac{Given mass}{Molecular mass}$ = $\frac{10^{-3} }{40}$ = $\frac{1}{(1000)(40)}$

                                              = $\frac{1}{40000}$ moles

Therefore, molarity = (1/40000 ÷ 250) × 1000 = 1/40 × 1/250

                                = 1/10000 =  $10^{-4}$ M

∴ Molarity of the NaOH solution is $10^{-4}$ M.

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