Calculate the weight of anhydrous AR sodium carbonate (Na2CO3) required to make 250.0mL of 0.0500 M solution?

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Answer 1 · 2018-04-29T20:09:40

This question is asking about how much of a compound (in terms of mass) we need to dissolve to make a solution of a given concentration.

Recall,

#"M" = "mol"/"L"#

Given,

#V = 0.250"L"#

#[Na_2CO_3] = 0.0500"M"#

Hence,

#"mol" = "L" * "M" = 1.25*10^-2"mol" * (106"g")/(Na_2CO_3) approx 1.33"g"#

of sodium carbonate are needed to make the solution requested.