Question

How many ml of 0.213-M Na3PO4 are required to deliver 66.4 mmol Na1+ ions?

Answer 1

Given :

Molarity of $Na_3PO_4$ , M = 0.213 M .

To Find :

How many ml of 0.213 M of $Na_3PO_4$ are required to deliver 66.4 mmol $Na^+$.

Solution :

Volume required  :

$V=[\text{ 66.4 mmol of }Na^+] + [\dfrac{\text{1 mm of }Na_3PO_4}{\text{3 mmol of }Na^+}]+[\dfrac{\text{1 ml }Na_3PO_4}{\text{0.213 mmol of }Na_3PO_4}]$

So ,

$V=\dfrac{66.4}{3\times 0.213}\ ml\\\\V=103.91\ ml$

Therefore , volume of $Na_3PO_4$ required is 103.91 ml .

Hence , this is the required solution .