Question

Moles of Na+ ions present in 500 cm cube of 0.25 M Na2SO4 solutions are:
(A) 0.25
(B) 0.125
(C) 0.375
(D) 0.5​

Answer 1

Given :

• Moles of $\tt{Na^{+} ions}$ present in=500cm³

• 1000ml of 0.25M $\tt{Na_{2}SO_{4}}$=0.25mol

To find :

• Moles of $\tt{Na^{+} \: ion}$=?

Solution :

As we know that,

$\bullet \sf \: 500 {cm}^{3} = 500ml$

$\\ \implies\sf{1000ml \: of \: 0.25M \: Na_{2}SO_{4} \mapsto 0.25ml }$

$\\ \implies\sf{500ml \: of \: 0.25M \: Na_{2}SO_{4} \mapsto \dfrac{0.25}{1000} \times 500=0.125mol}$

$\\ \implies\sf{1 \: mole \: Na_{2}SO_{4} \: \longrightarrow \: 2 \: mole \: of \: Na}$

$\\ \implies\sf{0.125 \: mole \: Na_{2}SO_{4} \longrightarrow 2 \times 0.125 =0.25 Mole \: Na}$

The number of moles of Na is 0.25 .

Answer 2

0.375 is right answer