Question

Calculate the number of grams of anhydrous ${Na}_2{CO}_3$ present in $250mL$ of $0.25M$ solution.

Answer 1

Volume of solution $=250$ $mL$

Molarity of solution $=0.25$

$M=\frac{No.ofmolesinsolute}{Volumeinlitres}$

$⟹0.25=\frac{No.ofmolesofN{a}_{2}C{O}_3}{250\times {10}^{-3}L}$

$⟹No.ofmoles=62.5\times {10}^{-3}=6.25\times {10}^{-2}$