Question

Sodium phosphate dissolves as follows: na3po4(s) → 3na (aq) po4-(aq). How many moles of na3po4 are required to make 1. 0 l of solution in which the na concentration is 0. 15 m?

Answer 1

The 0.05 moles of $Na_{3}PO_{4}$ required to make 1 L of solution in which the $Na$ concentration is 0. 15 M.

Calculation,

The Sodium phosphate dissolves as follows:

$Na_{3}PO_{4}(s)$ → $3Na^{+} (aq)+PO_{4}^{-} (aq)$

To make 1. L of solution in which the  $Na^{+$concentration is 0. 15 M

The morality  of $Na^{+$= 0.15 M = number of moles / volume in lit

The morality of $Na^{+$= 0.15 M = number of moles /1 Lit

number of moles of $Na^{+$ = 0.15 mole

The mole ratio is 1: 3

It means, the 1 mole of $Na_{3}PO_{4}(s)$ required to form 3 mole of $Na^{+$

So, to form 0.15 mole of $Na^{+$  = 1×0.15 / 3 moles of $Na_{3}PO_{4}(s)$required.

Hence, number of moles of $Na_{3}PO_{4}(s)$ required = 0.05 moles.

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