# If 25.00 mL of 1.04 M Na_2CO_3 is diluted to 0.500 L, what is the molarity of Na_2CO_3 in the...

## Question:

If {eq}25.00 \ mL {/eq} of {eq}1.04 \ M \ Na_2CO_3 {/eq} is diluted to {eq}0.500 \ L {/eq}, what is the molarity of {eq}Na_2CO_3 {/eq} in the diluted solution?

## Calculating Concentration:

The concentration of the diluted solution is calculated using the dilution process of a concentrated solution involved addition of water to obtain the diluted solution. Basically, this can be described by avertain equation given by {eq}M_1V_1 = M_2V_2 {/eq}, where {eq}M_1{/eq} is the initial concentration, {eq}V_1{/eq} is the initial volume, {eq}M_2{/eq} is the final concentration, {eq}V_2{/eq} is the initial volume.

The concentration of the diluted solution is calculated using

{eq}M_1V_1 = M_2V_2 {/eq}

where,

• {eq}M_1 = 1.04\ M {/eq} is the intial concentration,
• {eq}V_1 = 25.0 \ mL {/eq} is the intial volume
• {eq}M_2 {/eq} is the final concentration
• {eq}V_2 = 0.500\ = 500 \ mL {/eq} is the final volume.

Putting {eq}M_2 {/eq} on one side of the equation and substituting all given values. Therefore,

{eq}M_2 = \dfrac{M_1V_1}{V_2}\\ M_2 = \dfrac{(1.04\ M)(25.0\ mL)}{500\ mL}\\ \boxed{M_2 = 0.0520\ M} {/eq}